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itiiiirarp
Cornell University Library
QA 154.H37
Advanced algebra,
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MATHEMATICAL TEXTS
Edited by Percey P. Smith, Pii.C
Professor of Mathematics in the SheiHeld Scientific School
of Yale University
Elements of the Differential and Integral Calculus
(Kevised Edition)
By W. A. Granville, Ph.D., LL.D.
Elements of Analytic Geometry
By P. F. Smith and A. S. Gale, Ph.D.
New Analytic Geometry
By P. F. Smith and A. S. Gale, Ph.D.
Introduction to Analytic Geometry
By P. F. Smith and A. S. Gale, Ph.D.
Advanced Algebra
By H. E. Hawkes, Ph.D.
Text-Book on the Strength of Materials (Revised Edition)
By S. E. Slocum, Ph.D., and E. L. Hancock, M.Sc.
Problems in the Strength of Materials
By William Kent Shepard, Ph.D.
Plane and Spherical Trigonometry and Four-Place
Tables of Logarithms
By W. A. Granville, Ph.D., LL.D.
Plane and Spherical Trigonometry
By W. A. Granville, Ph.D., LL.D.
Plane Trigonometry and Four-Place Tables of Logarithms
By W. A. Granville, Ph.D., LL.D.
Four-Place Tables of Logarithms
By W. A. Granville, Ph.D., LL.D.
Theoretical Mechanics
By P. F. Smith and W. R. Longlet, Ph.D.
First Course in Algebra
By H. E. Hawkes, Ph.D., William A. Lubt, A.B.,
and Frank C. Touton, Ph.B.
Second Course in Algebra
By H. E. Hawkes, Ph.D., William A. Luby, A.B.,
and Frank C. Touton, Ph.B.
Complete School Algebra
By H. E. Hawkes, Ph.D., William A. Luby, A.B.,
and Frank C. Touton, Ph.B.
Elementary Analysis
By P. F. Smith and W. A. Granville, Ph.D., LL.D.
Plane Geometry
By William Betz, A.M. , and Harrison E.Webb, A.B.
ADVANCED ALG^^^A
BY
HERBEET E. HAWKES, Ph.D.
Assistant Pkopbssob of Mathematics in Yale Univebsitt
GINN & COMPANY
BOSTON • NEW YORK • CHICAGO ■ LONDON
H37
Copyright, 1906, by
H. E. HAWKES
ALL EIGHTS KESEKVED
613.1
GINN & COMPANY . PRO-
PRIETORS ■ BOSTON ■ U.S.A.
PEEFACE
This book is designed for use in secondary schools and in
short college courses. It aims to present in concise but clear
form the portions of algebra that are required for entrance to
the most exacting colleges and techidcal schools.
The chapters on algebra to quadratics are intended for a
review of the subject, and contain many points of view that
should be presented to a student after he has taken a first
course on those topics. Throughout the book the attention
is concentrated on subjects that are most vital, pedagogically
and practically, while topics that demand a knowledge of the
calculus for their complete comprehension (as midtiple roots,
and Sturm's theorem) or are more closely related to other por-
tions of mathematics (as theory of numbers, and series) have
been omitted.
The chapter on graphical representation has been intro-
duced early, in the beUef that the Ulumination which it affords
greatly enlivens the entire presentation of algebra. The dis-
cussion of the relation between pairs of linear equations and
pairs of straight lines is particularly suggestive.
In each chapter the discussion is directed toward a definite
result. The chapter on theory of equations aims to give a
simple and clear treatment of the method of obtaining the
real roots of an equation and tne theorems that lead to that
iv PREFACE
process. Similarly direct in its argument is the chapter on
determinants, its object being the solution of non-homogeneous
equations and the necessary evaluation of determinants.
I am under obligations to many friends and colleagues for
suggestions, but especially to Professor P. F. Smith, who has
read the book both in manuscript and proof and whose numer-
ous suggestions have been invaluable.
New Haven, Connecticut
August, 1905
CONTENTS
ALGEBRA TO QUADRATICS
CHAPTER I
FUHDAMENTAL OPERATIONS
SECTION PAGE
2. Addition 1
3. Subtraction 1
4. Zero 2
5. Multiplication 2
6. Division S
7. Division by Zero 4
8. Fundamental Operations 5
9. Practical Demand for Negative and Practional Numbers . . 5
10. Laws of Operation 5
11. Integral and Rational Expressions 6
12. Operations on Polynomials . . 6
13. Addition of Polynomials 7
?4. Subtraction of Polynomials . . 7
15. Parentheses 8
16. Multiplication 9
17. Multiplication of Monomials by Polynomials .... 10
18. Multiplication of Polynomials .... . . 10
19. Types of Multiplication 11
20. ' The Square of a Binomial 11
21. The Square of a Polynomial 12
22. The Cube of a Binomial 12
23. Division 12
24. Division of Monomials 13
25. Division of a Polynomial by a Monomial 13
26. Division of a Polynomial by a Polynomial 13
27. Types of Division 15
V
VI
CONTENTS
CHAPTER n
FACTORING
8IIOTIOI7
28. Statement of the Problem
29. Monomial Factors ....
30. Factoring by grouping Terms .
31. Factors of a Quadratic Trinomial
32. Factoring the Difference of Squares
33. Reduction to the Difference of Squares
34. Replacing a Parenthesis by a Letter
35. Factoring Binomials of the Form a" ± b"
36. Highest Common Factor ....
37. H.C.F. of Two Polynomials .
38. Euclid's Method of finding the H.C.F. .
39. Method of finding the H.C.F. of Two Polynomials
40. Least Common Multiple
41. Second Rule for finding the Least Common Multiple
TA.as
. 16
16
. 17
18
. 20
20
. 21
22
. 22
23
. 23
24
CHAPTER III
FRACTIONS
42. General Principles .... 27
43. Principle I 27
44. Principle II 27
45. Principle III 27
46. Reduction 27
47. Least Common Denominators of Several Fractions . . 28
48. Addition of Fractions 29
49. Subtraction of Fractions 29
50. Multiplication of Fractions 29
51. Division of Fractions 29
CHAPTER IV
EQUATIONS
52. Introduction 32
53. Identities and Equations of Condition 32
54. Linear Equations in One Variable 33
55. Solution of Problems 37
66. Linear Equations in Two Variables 40
57. Solution of a Pair of Equations 40
CONTENTS
VU
SXOTIOK PAGE
58. Independent Equations 41
59. Solution of a Pair of Simultaneous Linear Equations ... 42
60. Incompatible Equations 42
61. Rfeum^ 43
62. Solution of Problems involving Two Unknowns . . . .45
63. Solution of Linear Equations in Several Variables ... 47
CHAPTER V
RATIO AND PROPORTION
64. Ratio
65. Proportion ....
66. Theorems concerning Proportion
67. Tlieorem ....
68. Mean Proportion
49
49
49
50
50
CHAPTER VI
IRRATIONAL NUMBERS AND RADICALS
69. Existence of Irrational Numbers .
70. The Practical Necessity for Irrational Numbers
71. Extraction of Square Root of Polynomials .
72. Extraction of Square Root of Numbers ■ .
78. Approximation of Irrational Numbers .
74! Sequences
75. Operations on Irrational Numbere
76. Notation
77. Other Irrational Numbers
78. Reduction of a Radical to its Simplest Porm
79. Addition and Subtraction of Radicals
80. Multiplication and Division of Radicals .
81. Rationalization . ....
82. Solution of Equations involving Radicals
52
53
53
54
55
56
56
57
57
58
59
60
61
63
CHAJPTER VII
THEORY OF INDICES
83. Negative Exponents ... .... 66'
U^i. Fractional Exponents . 66
85. Further Assumptions .... .... 67
86. Theorem .67
87. Operations with Radical Polynomials 69
VIU
CONTENTS
QUADEATICS AND BEYOND
CHAPTER Vin
QUADRATIC EQUATIOITS
SECTION
88. Definition
89. Solution of Quadratic Equations .
90. Pure Quadratics
91. Solution of Quadratic Equations by Factoring
92. Solution of an Equation by Factoring
93. Quadratic Form . . • .
94. Problems solvable by Quadratic Equations
95. Theorems regarding Quadratic Equations .
96. Theorem
97. Theorem
98. Nature of the Roots of a Quadratic Equation
70
70
72
75
75
77
79
82
83
84
84
CHAPTER IX
GRAPHICAL REPRESENTATION
99. Representation of Points on a Line 87
100. Cartesian CoSrdinates 88
101. The Graph of an Equation 90
102. Restriction to Coordinates 91
103. Plotting Equations 91
104. Plotting Equations after Solution 93
105. Graph of the Linear Equation 94
106. Method of plotting a Line from its Equation .... 96
107. Solution of Linear Equations, and the Intersection of their Graphs 97
108. Graphs of Dependent Equations 99
109. Incompatible Equations 99
110. Graph of the Quadratic Equation 100
111. Form of the Graph of a Quadratic Equation .... 101
112. The Special Quadratic aa;^ + 6a; = 103
113. The Special Quadratic aa;^ + c = 104
114. Degeneration of the Quadratic Equation 104
115. Sum and Difference of Roots 106
116. Variation in Sign of a Quadratic 107
CONTENTS
CHAPTER X
SIMULTANEOUS QUADRATIC EQUATIONS IN TWO VARIABLES
SECTION PAGE
117. Solution of Simultaneous Quadratics . . . Ill
118. Solution by Substitution .... . . Ill
119. Number of Solutions ... 113
120. Solution when neither Equation is Linear . . . 114
121. Equivalence of Pairs of- Equations 120
122. Incompatible Equations 121
123. Graphical Representation of Simultaneous Quadratic Equations 122
124. Graphical Meaning of Homogeneous Equations . . . 123
CHAPTER XI
MATHEMATICAL INDUCTION
125. General Statement . . . .... 125
CHAPTER Xn
BINOMIAL THEOREM
126. Statement of the Binomial Theorem 128
127. Proof of the Binomial Theorem . .... 129
128. General Term .... .... 129
CHAPTER Xin
ARITHMETICAL PROGRESSION
129. Definitions . . 133
130. The Tith Term . . 133
131. The Sum of the Series 134
132. Arithmetical Means 134
CHAPTER XIV
GEOMETRICAL PROGRESSION
133. Definitions 137
134. The nth Term 137
135. The Sum of the Series . . 138
136. Geometrical Means 138
137. Infinite Series 140
X CONTENTS
ADVAlfCED ALGEBEA
CHAPTER XV
PERMUTATIONS Am> COMBHrATIONS
SECTION PAGE
138. Introduction 143
139. Permutations 144
140. Combinations 146
141. Circular Permutations 149
142. Theorem 150
CHAPTER XVI
COMPLEX NUMBERS
143. The Imaginary Unit " ... 152
144. Addition and Subtraction of Imaginary Numbers . . . 153
145. Multiplication and Division of Imaginaries 154
146. Complex Nunibers 155
147. Graphical Representation of Complex Numbers .... 165
148. Equality of Complex Numbers ... .... 155
149. Addition and Subtraction 156
150. Graphical Representation of Addition 156
151. Multiplication of Complex Numbers 157
152. Conjugate Complex Numbers 158
163. Division of Complex Numbers 158
164. Polar Representation 160
155. Multiplication in Polar Form 160
156. Powers of Numbers in Polar Form 161
157. Division in Polar Form 162
158. Roots of Complex Numbers 162
CHAPTER XVII
THEORT OF EQUATIONS
159. Equation of the nth Degree 166
160. Remainder Theorem 166
161. Synthetic Division 167
162. Proof of the Rule for Synthetic Division 169
163. Plotting of Equations 170
164. Extent of the Table of Values 171
CONTENTS xi
SECTION PAGK
165. Roots of an Equation 172
166. Number of Koots 172
167. Graphical Interpretation 174
168. Imaginary Roots 174
169. Graphical Interpretation of Imaginary Roots .... 175
170. Relation between Roots and CoefBolents 177
171. The General Term in the Binomial Expansion .... 178
172. Solution by Trial 178
173. Properties of Binomial Surds 179
174. Formation of Equations 180
175. To multiply the Roots by a Constant 183
176. Descartes' Rule of Signs 186
177. Negative Roots 189
178. Integral Roots .... 190
179. Rational Roots 190
180. Diminishing the Roots of an Equation . . . . 191
181. Graphical Interpretation of Decreasing Roots .... 193
182. Location Principle 194
183. Approximate Calculation of Roots by Horner's Method . . 195
184. Roots nearly Equal 200
CHAPTER XVni
DETERUINAHTS
185. Solution of Two Linear Equations 203
186. Solution of Three Linear Equations 204
187. Inversion 208
188. Development of the Determinant 208
189. Number of Terms 210
190. Development by Minors 210
191. Multiplication by a Constant 213
192. Interchange of Rows and Columns 213
193. Interchange of Rows or Columns 214
194. Identical Rows or Columns 215
195. Proof for Development by Minors 215
196. Sum of Determinants 216
197. Vanishing of a Determinant 217
198. Evaluation by Factoring ... .... 218
199. Practical Directions for evaluating Determinants . . . 219
200. Solution of Linear Equations 221
201. Solution of Homogeneous Lineai: Equations .... 223
xii CONTENTS
CHAPTER XIX
PARTIAL FRACTIONS
SEOTIOir PAGE
202. Introduction 225
203. Development when 0(iE) = has no Multiple Roots . . . 225
204. Development when ^ (a;) = has Imaginary Roots . . . 229
205. Development when 0(a:) = (k— a)" 232
206. General Case 238
CHAPTER XX
LOGARITHMS
207. Generalized Powers 235
208. Logarithms 236
209. Operations on Logarithms 237
210. Common System of Logarithms 239
211. Use of Tables 241
212. Interpolation 242
213. Antilogarithms 243
214. Cologarithins 245
215. Change of Base 247
216. Exponential Equations 251
217. Compound Interest 253
CHAPTER XXI
CONTINUED FRACTIONS
218. Definitions 256
219. Terminating Continued Fractions 256
220. Convergents 258
221. Recurring Continued Fractions 260
222. Expression of a Surd as a Recurring Continued Fraction . 263
223. Properties of Convergents 265
224. Limit of Error 267
CHAPTER XXn
INEQUALITIES
225. General Theorems 269
226. Conditional LinearTnequalities 271
227. Conditional Quadratic Inequalities 271
CONTENTS xiii
CHAPTER XXm
VAWATIOH
SECTION PA6B
228. General Principles . ,273
CHAPTER XXIV
PROBABILITY
''^229. Illustration 276
230. General Statement . . . . ■ 276
CHAPTER XXV
SCALES OF NOTATION
231. General Statement 279
232. Fundamental Operations ... .... 280
233. Change of Scale .... 281
234. Fractions 282
236. Duodecimals . . 284
ADVANCED ALGEBRA
ALGEBEA TO QUADEATICS
CHAPTEE I
FUNDAMENTAL OPERATIONS
1. It is assumed that the elementary operations and the mean-
ing of the usual symbols of algebra are familiar and do not
demand detailed treatment. In the following brief exposition
of the formal laws of algebra most of the proofs are omitted.
2. Addition. The process of adding two positive integers a and
6 consists in finding a number x such that
a + 6 = a;.
Por any two given positive integers a single sum x exists
which is itself a positive integer.
3. Subtraction. The process of subtracting the positive num-
ber h from the positive number a consists in finding a number x
such that
6 + a; = a. (1)
This number x is called the difference between a and h and is
denoted as follows :
a — 6 = K,
a being called the minuend and h the subtrahend.
If a > 6 and both are positive integers, then a single posi-
tive integer x exists which satisfies the condition expressed by
equation (1)
1
2 ALGEBRA TO QUADRATICS
li a Kb, then x is not a positive integer. In order that the pro-
cess of subtraction may be possible in this case also, we introduce
negative numbers which we symbolize by (—a ), (— J), etc. When
in the difference a — b,ais less than b, we define a — b = (— (b — a)).
The processes of addition and subtraction for the negative numbers
are defined as follows :
(-a) + (-b) = (-(a + b)).
(_ a) _(-&) = (-(« -5)).
i-a)-b=(-(a + b)).
a — ( — S) = a + S.
(-(-«))=«■*
4. Zero. If in equation (1), a = b, there is no positive or nega-
tive number which satisfies the equation. In order that in this
case also the equation may have a number satisfying it, we intro-
duce the number zero which is symbolized by and defined by
the equation
a + = a,
or a — a = 0.
The processes of addition and subtraction for this new number
zero are defined as follows, where a stands for either a positive or
a negative number
— a = — a.
± = 0.
5. Multiplication. The process of multiplying a by 5 consists
in finding a number x which satisfies the equation
a-b ^ X.
* The symbol for a positive integer might be "written (+ a), (+ 6), etc., consistently
with the notation for negative nvimbers. Since, however, no ambiguity results, Tre omit
the + sign. Since the laws of combining the + and — signs given in this and the following
paragraphs remove the necessity for the parentheses in the notation for the negative
number, we shaU omit them where no ambiguity results.
FUNDAMENTAL OPEEATIONS 3
When a and b are positive integers a; is a positive integer which
may be found hy adding a to itself b times. "WTien the numhers
to be mnltiplied are negative we have the following laws,
(— a) ■ (— J) = a • b,
(-a)-b = a-(-b) = -(a-b),
• a; = a ■ = 0, (1)
where a is a positive or negative number or zero.
These symbolical statements include the statement of the
following
Principle. A product of numbers is zero when and only when
one or more of the factors are zero.
This most important fact, which we shall use continually, assures
us that when we have a product of several numbers as
a-b-c-d = 6,
first, if e equals zero, it is certain that one or more of the num-
bers a, b, 0, or d are zero ; second, if one or more of the numbers
a, b, c, or d are zero, then e is also zero.
6. Division. The process of dividing a by ^8 consists in finding
a number x which satisfies the equation
x.l3 = a, (1)
where a and /3 are positive or negative integers, or a is 0.
When a occurs in the sequence of numbers
■■■-3p, -2p, -p, 0, A 2A 3A •••,
aj is a definite integer or 0, that is, it is a number such as we
, have previously considered. If a is not found in this series, but
is between two numbers of the series, then in order that in this
case the process may also be possible jre introduce the fraction
which we symbolize by a -s- )8 or - and which is defined by the
equation '^
4 ALGEBRA TO QUADRATICS
The operations for addition, subtraction, nmltiplieation, and divi-
sion of fractions are defined as follows :
^±t=
/3S
)
(2)
i8 8
)38'
(3)
/8 ■ 8
(4)
Further
properties
of fractions are the
following :
1 =
a 1
= « = !'
a
= — ) where 8 is any
munher,
(5)
— a
-P
a
(6)
The last two equations are expressed verbally as follows :
Both numerator and denominator of a fraction may be multi-
plied by any number without changing the value of the fraction.
Changing the sign of either numerator or denominator of a
fraction is equivalent to changing the sign of the fraction.
The laws of signs in multiplication given on p. 3 may now be
assumed to hold when the letters represent fractions as well as
integers. p ,•.-, p ,.
Thus for example — (t) " ~"(j)
ac
bd'
The positive or negative number a may be written in the
fractional form
a
T"
7. Division by zero. If in equation (1), § 6, /3 = 0, there is no
single number x which satisfies the equation, since by (1), § 5,
whatever value x might have, its product with zero would
be zero.
FUNDAMENTAL OPERATIONS 5
Thus division by zero is entirely excluded from algebraic pro-
cesses. Before a division can safely be performed one must be
assured tbat the divisor cannot vanish. In the equation
40 = 2-0,
if we should allow division of both sides of our equation by zero,
■we should be led to the absurd result 4 = 2.
8. Fundamental operations. The operations of addition, sub-
traction, multiplication, and division we call the four fundamental
operations. Any numbers that can be derived from the number 1
by means of the four fundamental operations we call rational num-
bers. Such numbers comprise air positive and negative integers
and such fractions as have integers for numerator and denominator.
Positive or negative integers are called integral numbers.
9. Practical demand for negative and fractional numbers.
In the preceding discussion negative numbers and fractions have
been introduced on account of the mathematical necessity for
them. They were needed to make the four fundamental opera-
tions always possible. That this mathematical necessity corre-
sponds to a practical necessity appears as soon as we attempt
to apply our four operations to practical affairs. Thus if on a
certain day the temperature is -f- 20° and the next day the mer-
cury falls 26°, in order to express the second temperature we
must subtract 26 from 20. If we had not introduced negative
numbers, this would be impossible and our mathematics would
be inapplicable to this and countless other everyday occurrences.
10. Laws of operation. All the numbers which we use in
algebra are subject to the following laws.
Commutative law of addition. This law asserts that the value
of the sum of two numbers does not depend on the order of
summation.
Symbolically expressed,
a + b = b + a,
where a and b represent any numbers such as we have presented
or shall hereafter introduce.
6 ALGEBRA TO QUADRATICS
Associative law of addition. This law asserts that the stun of
three mimbers does not depend on the -way in which the numbers
are grouped in performing the process of addition.
Symbolically expressed,
a + (b + c) = (a + b)+ e = a + b + c.
Commutative law of multiplication. This law asserts that the
value of the product of two numbers does not depend on the order
of multiplication.
Symbolically expressed,
a-b = b- a.
Associative law of multiplication. This law asserts that the
value of the product of three numbers does not depend on the way
in which the numbers are grouped in the process of multiplication.
Symbolically expressed,
(a-b") -0 = a- (b- c)= a-b- c.
Distributive law. This law asserts that the product of a single
number and the sum of two nuinbers is identical with the sum
of the products of the first number and the other two numbers
taken singly.
Symbolically expressed,
a' (b + c)= a-b + a- c.
All the above laws are readily seen to hold when more than three numbers are
involved.
11. Integral and rational expressions. A polynomial is inte-
gral when it may be expressed by a succession of literal terms, no
one of which contains any letter in the denominator.
Thus 4 j:^ — ajS — 2 i>;2 — I a + 1 is integral.
The quotient of two integral expressions is called rational.
„,^ a!2 - 2a; + 3 .
Thus = — IS rational.
x — 7
12. Operations on polynomials. We assume that the same
formal laws for the four fundamental operations enunciated in
§§ 2-6 and the laws given in § 10 hold whether the letters in the
symbolic statements represent numbers or polynomials.
FUNDAMENTAL OPERATIONS 7
In fact the literal expressions wliicli we use are in essence
nothing else than numerical expressions, since the letters are merely
symbols for numbers. When the letters are replaced by numbers,
the literal expressions reduce to numerical expressions for which
the previous laws have been explicitly given.
13. Addition of polynomials. For performiag this operation
we have the following
EuLE. Write the terms with the same literal part in a column.
Find the algebraic sum, of the terms in each column, and write
the results in succession with their proper signs.
When the polynomials reduce to monomials the same rule is to be observed.
EXERCISES
Add the following :
1. Za^lfi -Viob + &a?h - a; 4a6 - 2a262 _ UaSft + 9a;
2a26-a6-2a; %a'^l^-iah -6a.
Solution : 3 a^ft^ _ 2 a6 + 6 a^b - a
- 2 a262 + 4 a6 - 11 a^a + 9 a
- ab+ 2a?b-2a
8 ggftii - 4 gfe -6a
9a262-3a6- Sa^b
2. 21a-246-8c2; 16c + 176 + 6c2 - 20a ; 186-18c.
3. x^ - 6x^ - Sx - I ; 2x^ + 1 ; exfi + 1 X + 2 ; x^ - x' + X - 1.
4. 9(a + 6)-6(6 + c) + 7(a + c); 4(6 + c)-7(a + 6)-8(a + c);
(a + c)-{ar+b) + (b + c).
5. a2-4o6 + 62 + a + 6-2; 2a2 + 4a6 - 36^ _ 2a - 26 + 4;
3a2- 6a6-462 + 3a + 36-2; Ba" + 10a6 + 562 + a + 5.
14. Subtraction of polynomials. For Jjerforming this operar
tion we have the following <
EuLE. Write the subtrahend under the minuend so that terms
unth the same literal part are in the same column.
To each term of the minuend add the corresponding term of
the subtrahend, the sign of the latter having been changed.
It is generally preferable to imagine the signs of the subtrahend changed rather
than actually to write it with the changed signs.
8 ALGEBRA TO QUADRATICS
EXERCISES
1. From a262 - 3 a^ft + 8 a5 + 6 6 subtract 9 a?}fi - 6 o6 + 4 0^% + a.
Solution : a^fi" - 3 a^fe + 8 a6 + 6 &
Qam + iaV}- 6a6 +a
-8a262-7a26+ 14a6 + 66-a
2. From 6 ate — 4Lmn + 6x subtract Sjnn + 6 ax — Aabx.
3. From m + an + bq Subtract the sum of
cm + dn + {b — a)q and (a — 6) g — (a + d) n — cm.
4. From the sum of f a + ^^6 + Jc and - 6 — c — asubtract J6 - |c + ^a.
5. From the sum of 2a;2 _ 3a; + 4 and a;* — fa; - J subtracta;' — ix"
-3|s + 3J.
15. Parentheses. When it is desirable to consider as a single
symbol an expression uivolving several numbers or symbols for
numbers, the expression is inclosed in a parenthesis. This paren-
thesis may then be used in. operations as if it were a single number
or symbol, as in fact it is, excepting that the operations inside
the parenthesis may not yet have been carried out.
EuLE. When a single parenthesis is preceded hy a + sign
the parenthesis may he removed, the various terms retaining
the same sign.
When a single parenthesis is preceded hy a — sign the pareTV-
thesis may he removed, providing we change the signs of all the
terms inside the parenthesis.
When several parentheses occur in an expression we have the
following
EuLE. Remove the innermost parenthesis, changing the signs
of the terms inside if the sign preceding it is minus.
Simplify, if possible, the expression inside the new inner-most
parenthesis.
Repeat the process until all the parentheses are removed.
It is in geneial unwise to shorten the process hy carrying oat some of the steps
in one's head. The liability to error in such attempts more than offsets the gain
in time.
FUNDAMENTAL OPEEATIONS
EXERCISES
RemoTe parentheses from the following :
1. 6 - {9a - [26 + (4 a - 2 a - 6) - 6 6]}.
V^
Solution: 6 - {9a - [26 + (4a - 2o - 6) - 66]}
= 6 - {9a - [26 + (4o - 2a -i- 6) - 66]}
= 6-{9a-[26+(2a + 6)-66]}
= 6-[9a-(2 6 + 2a + 6-66)]
= 6 -[9a -(2a -36)]'
= 5-(9a-2a + 36)
= 5- (7a + 35)
= 6-7a-36
= - 7 a - 2 5.
2. _{__[_[_(_(_!))]]}.
3. a2 + 4 - {6 - [- (a2 - 6) + 1]}.
4- i{i-f[f-i(i-i-f-A) + f]-i}-
5. x2 - {y^ - [4s + 3(y - Qxiy - x)) + 9y(x- y)]}.
6. Find the value of a - {55 - [a - (3c - 36) + 2c - 3(a - 26 -c)]}
when a = — 3, 6 = 4, c = — 5.
.16. Multiplication. It is customary to write a- a = a' ;
a- a-a = a^; a-a- ■ ■a = a". We have then by the associative
n terras
law of nmltiplication, § 10,
or, in general, ^„^„ ^ ^„+.^ ^^
where m and. n are positive integers.
Furthermore,
(a")"" = »».»»...«•' = »»■■». (II)
V y '
TH terms
FinaUy, a' -h' = (a-i)\ (III)
The distinction between (a")"" and a"" should be noted carefully. Thus (28)"
= 82 = 64, while 28^ = 29 = 512.
Equation (I) asserts that the exponent in the product of two
powers of any expression is the sum of the exponents of the
factors. Hence we may multiply monomials as follows :
10 ALGEBRA TO QUADRATICS
EuLK. Write the product of the mumerical coefficients, followed
by all the letters that occur in the multiplier and multiplicand,
each having as its exponent the sum of the exponents of that
letter in the multiplier and multiplicand.
Example. 4 a,'b^°cd* ■ (- 16 0«&d') = - 64 oWicd".
17. Multiplication of monomials by polynomials. By the
distributive law, § 10, we can immediately formulate the
EuLE. Multiply each term of the polynomial by tJie monomial
and write the resulting terms in succession.
Example. 9 a'ft^ - 2 a6 + 4 aft^ - a + &«
Sa^b
. 27 o*68 - 6 am + 12 aW - 3 a^ft + Sa^b''
18. Multiplication of polynomials. If in the expression for the
distributive law, a(c + d)= ac + ad,
we replace a hj a + b, we have •
(a + b)(c + d)= ac + bc^ ad + bd,
which affords the
EuLE. Multiply the multiplicand by each term of the multi-
plier in turn, and write the partial products in succession.
To test the accuracy of the result assume some convenient numer- '
ical value for each letter, and find the corresponding numerical
value of multiplier, multiplicand, and product. The latter should
be the product of the two former.
EXERCISES
1. Multiply and check the following :
(a) 2 a^ + a6 + 4 &2 + 6 and a—b + ab.
Check :
Solution:
Let a = 6 = 1
2o2+ 06+462 + 6
= 8
a — b -\- ab
= 1
2 as + a26 + 4 ab^ + ab
-ia^b- c*2 -463-
-62
+ a62
+ 2aS6 + a=62 + 4a68
2a8 - a25 + 4a62 + a6 - 468 - 62 + 2a86 + a^62 4- 4<;^ = 8
FUNDAMENTAL OPERATIONS H
(b) 6abx^a,ni4a^b'x.
(0) ^^and -93^^22.
o
(d) 3 a^x -iW and 6 a'cx.
(e) x^" + y^'> + x!^ and a;" — yK
(f ) a2 + a6 + 62 and a^ + ac + A
(g) a!»2/™-2, ir»-'y"+*, and x^'^'^-'K
(h) KP-s + xP-^ + 1 and sS _ -^a _ l.
(i) 8 a?bc, I at^, - 7 62, - — a'c* and -.
4 ' ' 14 ' 6
(j) aa^ - 2 a?3? -x + ia and - a: + 2 a.
(k) xf+'> + a;2° + a:^!' + a;8»-6 and a;"-' — 1.
(1) 15a^ - lla;» + 6a;2 + 2x - 1 and - 3a;2 - 1.
(m) 4a^ - Sxy' + Ja;V - ^x^y ~x — y and — 42a;j/.
2. Expand (a; + y)K
3. Expand and simplify
(a;2 + 2,2 + 22)2 _ (a; + 2/ + z)(a; + y- 2)(a; + z - y)(y + z - ^.^^
19. Types of multiplication. The following types of multi-
plication should be so familiar as merely to require inspection of
the factors in order to write the product.
Rule. The product of the sum, and difference of two terms is
equal to the square of the terms with like signs minus the square
of the terms which have unlike signs.
Examples. (a - 6) (a + 6) = «« - 62. (1^-1 a 'b^^^
(4 a;2 - 3 y2) (4ij;2 + 3 2,2) _ 16 a^ _ 9 yi.
20. The square of a binomial. This process is performed as
follows :
EuLE. The square of a linomial, or expression in two terms,
is equal to the sum of the squares of the two terms plus twice
their product.
Examples. (a; + yY = v!^ + ifl + i xy.
(2o - 36)2 = 4a2 + 952- i^ab.
12
ALGEBRA TO QUADRATICS
21. The square of a polynomial. This process is performed
as follows :
EuLE, The square of any polynomial is equal to the sum
of the squares of the terms plus twice the product of each term
by each term that follows it in the polynomial.
Example, (o + 6 + c)2 = a^ + 6^ + c" + 2a6 + 2ac + 26c.
22. The cube of a binomial. This process is performed as
f ollo-ws :
EuLE. The eube of any binomial is given by the following
expression : ^^ + j)a = „a + 3 ^.j + 3 ^j. + 5..
EXERCISES
Perform the following processes by inspection.
1. (a-6 + c)2.
3. (2 a;'— 1-1)2.
5. (2 a-— '-1)8.
7. (x2-2a; + l)2.
9. (2»-2&-c)2.
11. (ai'-3-6!'+3)2.
13. (xp — y<i) {xP + y«).
15. (Sx + 2y)(3x-2y).
17. (-3s2y + Jz2)(3x2y + ^z2).
19
2.
4.
6.
8.
10.
12.
14.
16.
18.
20.
a* - 66)2
a2 - 62)3.
[X" - 2/2 + Z2)2.
■a;8_2s-l)2.
-6x^ + ixy^)«.
-6a2y + 4sj^2)2.
-3aa;2 + 2aa;-6)2.
-4-6a26)(-4 + 6a2i
(-r-
23. Division. By the definition of division in § 6, we have
-.2 „a _6
or, in general,
« 2 «
— r J a2 = —
a a
. „ a"
> a' = —
2'
(1)
where n and m are positive integers and n> m.
If re = m, we preserve the same principle and write ,
a"
FUNDAMENTAL OPERATIONS 13
24. Di-vision of monomials. Keeping in mind the rule of signs
for division given in § 6, we have the following
EuLE. Divide the numerical coefficient of the dividend ly
that of the divisor for the numerical coefficient of the quotient,
keeping in mind the rule of signs for division.
Write the literal part of the dividend over that of the
divisor in the form of a fraction, and perform on each pair
of letters occurring in both numerator and denominator the
process of division as defined by equation (I) in the preceding
paragraph. .
ExAMPLi. Divide 12 a*6"c2(Z by - 6 a^ic'd'.
12a«6"cgd _ 2 6M
- 6 a'ftc^d^ ~ a^cP '
25. Division of a polynomial by a monomial. This process is
performed as follows :
EuLE. Divide each term of the polynomial by the monomial
and write the partial quotients in succession.
Example. Divide 8 aV:fi — 12 a^ft^ i,y 2 o'6». *
8^ _ 12aW _ 4&S _ 6c^
26. Division of a polynomial by a polynomial. This process
is performed as follows :
EuLE. Arrange both dividend and divisor in descending
powers of some common letter (called the letter of arrangement)-
Divide the first term of the dividend by the first term of the
divisor for the first term of the quotient.
Multiply the divisor by this first term of the quotient and
subtract the product from the dividend.
Divide the first term of this remainder by the first term of the
divisor for the second term of the quotient, and proceed as
before until the remainder vanishes or is of lower degree in the
letter of arrangement than the divisor.
14 ALGEBRA TO QUADRATICS
When the last remainder vanishes the dividend is exactly divisible by the
divisor. This fact may be expressed as follows:
dividend _ ...
—n—. = qnotient.
divisor
When the last remainder does not vanish we may express the result of division
dividend _ ... . , rema inder
,. . = quotient H zr-.
divisor divisor
The coefficients in the quotient will be rational numbers if those in both divi-
dend and divisor are rational.
EXERCISES
Divide and check the following :
1. 8a' + 6a'^b + 9ab^ + 9i!'by ia + b.
Solution : 4a + 6| 8 a« + Ga^b + 9ab^ + 9 b' \2a^ + ab + 2b'
8 gg + 2 g^i)
4 a2& + 9 a62
4 aV) + of
8g62 + 96s
8g5' + 26»
7 6=
76'
Result: 2a= + g6 + 262 +
4g + 6
Check : Let = 6=1. Dividend = 32, divisor = 5, quotient = 6|.
32 -H 6 = 6|.
2. a;i2 - 2/12 by a;8 -y«. 3. 2!c2 _ 6a; + 2 by a; - 2.
4. x" — yi2 Tcj a;4 _ yi_ 5. 3.6 _ yS by x^ + xy + y^.
6. gs - a2 + 2 byg + 1. 7. - QZxhjH^ by - 9a;%'z.
8. a;2-x-30bya; + 5. 9. 4g26 - 6g62 -2g by -2a.
10. 16 a26«cii by - 2 amc\ 11. \x^ -^x- ^\)y l\x + ■^.
12. ax^ + (a2 — 6) a; - g6 by a; + g.
13. ax" + 6a;»-i + ca!»-2 — cke'-n by xK
14. 16 a^aiV _ 8 gaV - 4 a;*^ by - ^xy.
15. 18g!'63 + 6gp + 269+3- 9gp + i6«by 3ai'6».
16. g2-2o6-4c2 + 86c-362by a-2c + 6.
17. k' — (g + 6 + c) a;2 + (g5 + gc + 6c) x — a5c by k — a.
18. 2ya_6a!2 + J^a!2/ + J^x-^2/ + lby2s + 4y_|.
\9. x'-'ix'^ -x^ + j^ + ixy -y -y^ + 2\>jx-y + 1.
20. 3a» + Qx'h/ + 9x2 + 2x3^2 + 5j/» + 2y + 6y2 + sby x + 2y + 3.
FUNDAMENTAL OPERATIONS 15
27. Types of division. Tte following types of division, vhicli
may be verified by the rule just given for any particular integral
value of n, should be so familiar that they may be performed by
inspection.
((i2» — 62») -=- (ffi» ± b") = a" qi b". (I)
(a- + &») ^{a + b) = a"-^ - a—'b + a'^-W + 6"-^ (II)
where n is odd.
(a" _ b') -i-(a-b) = a"-i + a''-^b + a'^-W -{ h S"-', (III)
where n is odd or even.
EXERCISES
Give by inspection the results of the following divisions.
1. o6 - 1 by a - 1. 2. a6 + 1 by a + 1.
3. a' + 128 by a; + 2. 4. x^ + y^hjx + y.
5. ifi — y" by x^ + yK 6. x* — y* hj x — y.
7. a;8 - y' by a;* - y*. 8. d'"' — lhj a — 1.
9. a2»+i-lby a-1. 10. 27o9 + Sft'by 3a« + 26.
11. 8a;3-27by 2a:-3. 12. 4a2 - 2566' by 2a + 166*.
13. 16 a* - 256 by 4a2 + 16. 14. 27a" - 64612 by 3a* - 46*.
CHAPTEE II
FACTORING
28. Statement of the problem. The operation of division con-
sists in finding the quotient "when the dividend and divisor are
given. The product of the quotient and the divisor is the divi-
dend, and the quotient and the divisor are the factors of the
dividend. Thus the process of division consists in finding a
second factor of a given expression when one factor is given.
The process of factoring consists in finding all the factors of
a polynomial when no one of them is given. This operation is in
essence the reverse of the operation of multiplication. We shall
be concerned only with those factors that have rational coeflcients.
29. Monomial factors. By the distributive law, § 10,
ab + ac = a(b + e).
This affords immediately the
EuLE. Write the largest monomial factor which occurs in every
term outside a parenthesis which includes the algebraic sum of
the remaining factors of the various terms.
EXERCISES
Factor the following :
1. 6 aP'h^c + 9a66c4 - 15 a*6c'.
Solution : 6 a^We + 9 06%* - 15 a'^hc' = 3 a6c (2 afis + 3 66c« - 5 a»c«).
2. lianx — 21 bnx — In.
3. 121 a^ftSfi - 22 a'bc^ + 11 ab^c'.
4. 5 xV _ 10 x^y^ — 6 xV _ 15 ^fiyn,
5. 21 a6n + 6 o6%2 - 18 a^bn!' + 15 a^^n.
6. 10 db^cmx — 5 alficy + 5 ab^cz — 15 ab(?m\
7. 7 aHhj^ - 49 anV + 14 axyH^ - 21 aHy^.
8. 45 a*62c8<i - 9 ab>'(fld!' + 27 aPbc^d? - 117 aVficdK
16
FACTORING 17
30. Factoring by grouping terms. If in the expression for
the distributive law, „„ , i„ / , i,\„
' ac + 00 = (a + 0) c,
we replace c\>j c -\- d,
we have a(e + d) -\- b (o ■{- d) = ac -{- ad + he + hd.
We may then factor the right-hand member as follows :
ao + ad-\-bc + bd= a(e + d)-\-b(o + d) = (a, + b)(c + d).
This affords the
EuLE. Factor out any monomial expression that is common to
each term of the polynomial.
Arrange the terms of the polynomial to he factored in groups
of two or more terms each, such that in each group a monomial
factor may ie taken outside a parenthesis which in each case
contains the same expression.
Write the algebraic sum, of the monomial factors that occur
outside the various parentheses for one factor, and the expres-
sion inside the parentheses for the other factor.
EXERCISES
Eactor the following :
1. 4aS6-6a262_4a4 + 6a58.
Solution : 4 a'S - 6 a^b^ - 4 a* + 6 afts
= 2a(2a26-3a62_2a8 + 368)
= 2a(2a26 -2a? - ZaJfi + 368)
= 2a[2a2(6-o) + 362(6-a)]
= 2a(6-a)(2a2 + 362).
Solution: a;2 - (3a + 46)a; + 12a6
= a;2 - 3 CKC - 4 to + 12 a6-
= a; (a — 3 a) — 4 6 (a; — 3 a)
= (x-46)(a;-3a).
Z. 2ax-Zhy -2bx + %ay. 4. 56a2 - 40a6+ 63ac - 456c.
5. afz^ — My' — Ifix^ + aPyK 6. 91 x^ — 112mx + 65nx - 80 mn.
7. ax — bx + ex + ay — by + cy. 8. 2 aa; — 6a; — c5 + 2 a5 + 2 oc — 5*.
18 ALGEBRA TO QUADRATICS
9. 2a;»-3a;» + 2a;>-3. _ 10. a;' + a;' + a; + 1.
11.' acx^ - bcx + adz - bd. 12. 26a; - »« -46 + 2a!.
13. 3a;8 - 12s»2/a _ 4^= + 1. 14. 4x2 _ (s + b)x + 26.
15. a;* - (4m + 9n)a;2 + 36 mn. 16. x* - (2ot + 3n)x2 + 6mn.
17. a2a;-ac+o6y-o62x-682/+c62. 18. 18 a' - 2 aci6» - 9 0^6 + c^ft'.
19. 2aa;-6a2/ + a-26x + 56y-5. 20. 2ax-ay-25x+4cx-2cj/+52/.
21. 2 o2a;6 + 4 a^x* + 2 a^s" + 4 a".
22. 8x2_ 2ox-12xz + 3az + 4sy-ay.
31. Factors of a quadratic trinomial. In tliis case we cannot
factor by grouping terms immediately, as that method is inappli-
cable to a polynomial of less than four terms. We observe, how-
ever, that in the product of two binomial expressions,
(mx + n) (jpx -f- 2') = Tnpx^ + (mq + np)x -f- nq,
the coeflElcient of a: is the sum of two expressions mq and np,
whose product is equal to the product of the coefficient of x^ and
the last term, that is,
mq ■ np = mp ■ nq.
Thus, to factor the right-hand member of this equation, we may
remove the parenthesis from the term in x and use the principle
of grouping terms. Thus
mpx" + (mq + np) x + nq
= mpx^ + m,qx + npx +- nq
= mx (px + q)+ n (px + q)
= (mx + n) (px -f- q).
This affords the
EuLE. Write the trinomial in order of descending powers of
X (or the letter in which the expression is quadratic).
Multiply the coefficient of a^ ly the term not involving x, and
find two factors of this product whose algebraic sum 'is the
coefficient of x.
Replace the coefficient of x hy this sum and factor by group-
ing terms.
FACTORING 19
Factoring a perfect square is evidently a special case under this method.
Thus factor x^ + Qx+9.
1-9=9. 3 + 3 = 6.
a;2 + (3 + 3)a; + 9
= a;2 + 3a; + 3a; + 9
= !i;(x + 3) + 3(a; + 3)
= (a; + 3) (s + 3)
= {z + 3)K
One will usually recognize when a trinomial is a perfect square, in which
case the factors may be written down by inspection.
EXERCISES
Factor the following :
1. 3x^ + 8x + i.
Solution : 3 • 4 = 12. 6 + 2 = 8.
3a;2 + 8a; + 4
= 3a;2 + (6 + 2)x + i
= 3a;2 + 6a; + 2a; + 4
= 3a;(a; + 2) + 2(a; + 2)
= (3 a; + 2) (a + 2).
2. 8x2 -14 6a; + 3 62.
Solution: 8-362 = 2452. - 26 - 125 = - 146.
8a;2- 146a! + 362
= 8a;2-(25 + 126)x + 352
= 8a;2- 12 5a;- 26a; + 362
= 4a;(2a;-36) -6(2a;-36)
= (4a; -6) (2 a; -3 6).
3. 28a;2-3x-40.
Solution : 28 • (- 40) = - 1120.
The factors of 1120 must be factors of 28 and 40. We seek two factors
of 1120, one of which exceeds the other by 3. We note that since 40 exceeds
28 by more than 3, one factor must be greater and the other less than 28 and
40 respectively.
Since 4 - 7 = 28 and 5 • 8 = 40,
we try 5 - 7= 35 and 4 - 8 = 32,
which are the required factors of 1120.
28a;2-3a;-40
= 28a;2-(35-32)a;-40
= 28a;2-35x + 32a;-40
= 7a;(4a;-5) + 8(4a;-6)
= (7a; + 8)(4x-5).
20 ALGEBRA TO QUADRATICS
4. x2 - 6a; + 9. 5. 2x2 + a: _ 6.
6. 2x2 _x- 6. 7. 2x2 + a; -91.
8. x2 + X - 182. 9. 9x2 - 2x - 7.
10. 2x2 + 5a; + 3. 11. x2 - llx + 18.
12. 3x2-10x-8. 13. 6x2 + 17x + 7.
14. 15x2 + 4x- 3. 15. 7j,2_42,_ii.
16. o2-6a5 + 962. 17. 5x2 - 12x + 4.
18. 18x2-73x + 4. 19. 27x2 + 3x- 2.
20. 24x2 - Six - 15. 21. 21x2 - Six + 4.
22. 12x2 + 60x - 72. 23. x* - 3ox2 + 2a2.
24. 9x2-18ax-7a2. 25. 10x2-6Sx-lS.
26. xV - 12x2y8 + 36. 27. 4x21" - IBxP - 81.
28. 2x8 - 176x2 + 8&2x. 29.. 4a2 + 12a6 + 9&2.
30. 5a2x2-2a5x-762. 31. lOx* - 15a2x3 - 100x2ai.
32. 4 02"" + 16 a^b' + 16 62« 33. 4 a^xf'y* - 20 obxyH + 25 hH^.
32. Factoring the difference of squares. Under tlie method of
the preceding paragraph we may factor the difference of squares.
Thus to factor x^ — b^ "we observe that the product of the
coefB.eient of x^ and the constant term is
1 . (- &2) = _ S2_
Since the coefficient of x is zero, "we have
- 6 + & = 0.
Hence (x + b) (x — V) =■ x^ ~ b\
EuLE. Extract the square root of each term.
The sum of these square roots iff one factor, and their differ-
ence is the other.
Example. Factor 9 a2x»|/* — 16 6'c2.
9 a2x V - 16 68c2 = (3 axV + 4 ftic) (3 ox^ _ 4 6*c) .
33. Reduction to the difference of squares. The preceding
method may be used when the expression to be factored becomes
a perfect square by the addition of the square of some expression.
FACTORING 21
EXERCISES
Pactor the following :
1. a* + 46*.
Solution : a* + 4 6* = a* + 4 aW + 4 6* - 4 am
= (a2 + 2 62)2 _ 4 gib^
= iofl + 2 62- 2o6)(a2 + 262 + 2a6).
2. 1-a*. 3. a* + 4.
4. a;' — a;. 5. ofiy^ + ix\
6. 4x4 + 2/*. 7. 4a2_2562.
8. x* + x2 + l. 9. 16a26*-a;*.
10. X* + 9a;2 + 81. II. ia^p - ^b'^c'^i.
12. a2»'+3 - 16 0^6*. 13. a;*» + a;2» + 1.
14. 36a;2j,*a»- 49^2^16 15. cc* - 13a;2 + 36.
16. my* + 16 mx* - 12 )jia;"2/2. 17. 9a^ + 8 x^^ + iy*.
34. Replacing a parenthesis by a letter. Any of the preceding methods
may be applied when a polynomial appears in place of a letter in the expres-
sion to be factored. It is frequently desirable for simplicity to replace such
a polynomial by a letter, and in the final result to restore the polynomial.
EXERCISES
Factor the following :
1. 2aa;2-26a;2- 6aa; + 65x-8a + 86.
Solution : 2 aa;2 _ 2 5a;2 - 6 ax + 6 5a; - 8 a + 8 6
= 2(a-6)s2- 6(a-6)x-8(a-6)
= (a -6) (2x2 -6.x -8)
= 2(a-6)(x2-3x-4)
= 2 (a - 6) (X - 4) (X + 1).
In this example the factor (a — 6) might have heen replaced by a letter.
2. o2 + 62 _ c2 - 9 - 2 a6 + 6 c.
Solution: a2 + 62 - c2 - 9 - 2a5 + 6c
= a2 - 2a6 + 62 - (c2 - 6c + 9)
= (a - 6)2 - (c - 3)2
= (a - 6 + c - 3) (a - 6 - c + 3).
3. (3x - 2/) (2a +p) - (3x ~y)(a- q).
4. (4a-66)(3m-2p) + (a+ 56)(8m-2p).
5. (7a-3y)(5c-2(J)-(6a-2j/)(5c-2d).
6. (X - y) (3a + 46) - (4a - .56) (z-y)-{x- y) (2a - 85).
22 ALGEBRA TO QUADRATICS
7. 6(x + y)^-ll{x + y)-1.
8. 4a2-12oft + 9&2- s2-2a;-l.
9. x^cfi + 2xH + a;2 - a2 - 2a - 1.
10. ax^ + 6ax + 9a- bx^ - 66a; - 96.
11. 4(a - 5)2 - 5(a2 - 6^) - 21(a + 6)2.
12. 5 (s + !/)2 - 12 (x2 - j/2) + 4 (X - y)2.
13. a262x2 - a262 - 2 a5a;2 + 2 a6 + x^ _ i.
14. (x - 2y) (2a - 36) - (96 - 10) (x - 2y).
35. Factoring binomials of the form a" ± 6". By § 27,
a- - 6» = (a - b) (a"-' + a"-^b + a"-'b^ -\ \- S""').
a» + 6» = (a + b) (a"-^ - aP-'^b + aP-^V h *""')'
where n is odd.
One can factor by inspection any binomial of the given form
by reference to these equations.
EXERCISES
Factor the following :
1. x^ — y\
Solution : x^ — y^ ■= {x' — y') (x' + y')
= (a;2 + xy + y^) (x - y) (a^ -xy + y") \x + y).
2. x6 + 125. 3. a;' - 1.
4. a;i2 — 2/12. 5. x^ — y\
6. x" - 2/18. 7. x" - 2/i«.
8. a^xs + aS. 9. x* - aV.
10. 216 a + a*. 11. as* - 16 a.
12. 3 a' - 96 65a2. 13. 27 xV + 64 2/8.
14. 27 x52/' + x22/4. 15. 16 a''68 - 81 ci«d8.
36. Highest common factor. An expression that is not further
divisible into factors with rational coefficients is called prime.
If two polynomials have the same expression as a factor, this
expression is said to be their common factor.
The product of , the common prime factors of two polynomials
is called their highest common factor, or H.C.F.
The same common prime factor may occur more than once. Thus {x — 1)2 (x+1)
and (a; - 1)2 (a; - 2)2 have (a; - 1)2 as their H.C.F.
FACTOEING 23
37. H.C.F. of two polynomials. The process of finding the
H.C.F. is performed as follows :
EuLE. Factor the polynomials. The product of the common
prime factors is their H. G.F.
EXERCISES
Find the H.C.r. of the following:
1. 4 a62a;4 _ 8 ahH'^ + 4 ab^ and 6 oftx^ + 12 abx + 6 a6.
Solution : 4 aft^a;* - 8 06^2 _!_ 4 abi
= 4a&2(a;*-2a;2 + i)
= 4a62(a;-l)2(a; + l)2.
6o6a;2 + i2a6a; + 6a6
= 6a6(a;2 + 2a; + l)
= 6a6(a; + l)2.
The H.C.r. is then 2ab(x + 1)2.
2. a;^ — y^ and a;^ — y^.
3. a;3 + a;2 - 12a! and a;2 + 5a; + 4.
4. 9mx2 — 6 ma; + m and 9na;2 — n.
5. 6a;- 4x2 + 2aa;-3oand9-4a;2.
6. 12(i2-36a6 + 27 62and8a2-1862.
7. 3a%; - 6a5a; + Sft^a; &TX&ia^ - 46V
8. 2 a; - 46 - a;2 - 2 6a; and 4a; - 5a;2 - 6.
9. 6a;S -7oa!;2-20a2a;and3a;2 + aa;-4a2.
38. Euclid's method of finding the H.C.F. When one is unable to factor
the polynomials whose H.C.F. is sought, the problem may nevertheless be
solved by use of a method which in essence dates from Euclid (300 b.c).
The validity of this process depends on the following
Pkinoiple. If a polynomial has a certain factor, any multiple of it has the
same factor.
Let x" + Ax''-'- + Bx"-^ H \- K
and a?" + ax™-! + bx"'-^ + ■ ■■ +1
be represented by F and G respectively. The letters A,B,---,K and a, b,
■ ■ -,1 represent integers, and m, the degree of G, is no greater than n, the
degree of F. We seek a method of finding the H.C.F. of F and G if any
exists. Call Q the 6[uotient obtained by dividing F by G, and call B the
remainder. Then (§ 26)
F=QG + B, (1)
24 ALGEBRA TO QUADRATICS
where the degree of E in x is not so great as that of G. Now whatever the
H.C.F. of y and G may be, it must also be the H.C.F. of 6 andiJ. For since
F- QQ = B,
the H.C.F. of F and Q must be a factor of the left-hand member, and hence
a factor of B, which is equal to that member. Also every factor common
to G and B must be contained in F, for any factor of G and iJ is a factor of
the right-hand member of (1), and hence of F.
Thus our problem is reduced to finding the H. C. F. of G and B. Let Qi and
JJi be respectively the quotient and remainder obtained in dividing G by R.
Then G=QiE + iJi,
where the degree of JRi in a; is not as great as that of B. By reasoning simi-
lar to that just employed we see that the H.C.F. of G and B is also the
H.C.F. of B and Bj,. Continue this process of division.
Let B = Qa^x -I- -^2,
Bi = QsBx + JBg-
until, say in S* = Qt+sBi+i -t- iJ*+2,
either JBj is exactly divisible by Bk +i (i. e. Ej + 2 = 0) , or JJi + 2 does not con-
tain X. This alternative must arise since the degrees in x of the successive
remainders B, Bi, B^, • ■ • are continually diminishing, and hence either the
remainder must finally vanish or cease to contain K. Suppose iJi + 2=0. Then
the H.C.F. of iJi and iJi + 1 isiJi+i itself, which must, by the reasoning given
above, be also the H.C.F. of F and G. IfBt + i does not contain a;, then the
H.C.F. of F and G, which must also be a factor of iJi + 2, can contain no x,
and must therefore be a constant.
Thus F and G have no common factor involving x.
This process is valid if the coefficients of F and G are rational expressions in
any letters other than x.
39. Method of finding the H.C.F. of two polynomials. The above dis-
cussion we may express in the following
EuLE. Divide the polynomial of higher degree (if the degrees of the polyno-
mials are unequal) by the other, and if there is a remainder, divide the divisor
by it; if there is a remainder in this process, divide the previous remainder
by it, and so on until either there is no remainder or it does not contain the letter
of arrangement. If tliere is no remainder in tlie last division, the last divisor is
the H. C.F. If the last remainder does not contain the letter of arrangement,
then the polynomials have no common factor involving that letter.
In the application of this rule any divisor or remainder may be multiplied or
divided by any expression not involving the letter of arrangement without affect-
ing the H.C.F.
FACTORING
25
EXERCISES
Find the H.C.r. of the following :
1. 2x* + 2x'-x^-2x-l and x* + x' + ix + i.
Solution :
Multiply by -
lJ2x* + 2a;3-a2_ 2a5-l |x*+ x'+ 4a; + 4
2x* + 2x' + 8a: + 8
1, -ci;a-10a;-9
_^+10£+9Ja;*+ x'+ ix + i \x^-9x+81
x^ +Wx^+ 9x^
- 93?- 9x^+ ix
- 9a;8-90a;2- 81a;
Divide by - 725,
81a;''+ 85!i;+4
81a''+810a;+729
-725a;-725
a; + 9| a:' + 10a;+ 9 |a: + l
x^+ X
9a;+ 9
9a: + 9
Thus the H.C.r. is a; + 1.
This process may be performed in the following more compact form.
x^-9x+ 81
2
2x4 + 2a;s-
-a;2- 2a; -1
a;*+ x'+ ix + 4
2a;* + 2a;8
+ 8a; + 8
a;* + 10a;3+ 9x^
-1
-a;2-10a;-9
- 9a;3- 9a;2+ 4a;
s + 9
a;2 + 10x + 9
X^+ X
- 9a;3- 90a;2- 81a;
81a;2+ 85a; + 4
9a; + 9
9x + 9
81a;2 + 810a; +729
- 725 a; - 725
X +1
•725
Result : a; + 1.
2. a;2 + 6a; - 7 and a;' - 39a; + 70.
3. x^ — x*'-x + l and Sa;* - ix^ — 1.
4. x' + 2a;2 + 9 and - l,x^ - lla;" + 15x + 9.
5. a;8 - 2x2 - 15x + 36 and 3x2 - 4x - 15.
6. X* - 3x3 + x2 + 3x - 2 and 4x' - 9x2 + 2x + 3.
7. 4x» - 18x2 + 19x - 3 and 2x* - 12xS + 19x2 _ 6x + 9.
8. X* + 4xs - 22x2 - 4x + 21 and x* + lOxS + 20x2 - lOx - 21.
9. 6 a*x' - 9 a'x'y - 10 a^ xy2 + 15 ay^ and 10 a^xV _ 15 a*x^y' + 8 a'x'y*
- 12 o2xV-
26 ALGEBRA TO QUADRATICS
40. Least common multiple. The least common multiple of two
or more polynomials is the polynomial of least degree that con-
tains them as factors. We may find the least common multiple
of several polynomials by the following
EuLE. Multiply together all the factors of the various poly-
nomials, giving to each factor the greatest exponent with which it
appears in any of the polynomials.
41. Second rule for finding the least common multiple. When
only two polynomials are considered the previous rule is evidently
equivalent to the following
EuLE. Multiply the polynomials together and divide the
product hy their highest common factor.
EXERCISES
Find the least common multiple of the following :
1. s2 _ j/2^5;2 J^ay — ax — xy, and x"^ —2xy + y^.
Solution : x^ -y^ = {x -y)(x + y).
x^ + ay-ax — xy=:{x-y){x — a).
x^ - 2xy + y^ = ix - yf.
Thus the L.C.M. = (x - y)^ (x + y)(x- a).
2. ia?bc, 6a62, and 12 A
3. 9a;2/2, 6a;V, and Zxy^z''.
4. (X + 1) (x2 - 1) and x^ - 1.
5. x* + 4x22/2 and x'^ + 2y^-'2y.
6. 4x2 _ 92,2 and 4x2 -Vixy + 9y^.
7. a;2 _ 4 X + 3, x2 — 1, and x^ — ax — x + a.
8. X - 1, 2x2 - 5x - 3, and 2x8 _ 7a;2 + 2x + 3.
9. x8 - 9x2 + 26x - 24 and x' - lOx^ + 31 x - 30.
10. 2x2 - 3x - 9, x2 — 6x + 9, and 3x2 - 9x - 6x + 35.
CHAPTEE III
FRACTIONS
42. General principles. The symbolic statements of the rules
for the addition, subtraction, multiplication, and division of alge-
braic fractions are the same as the statements of the correspond-
ing operations on numerical fractions given in (2), (3), and (4),
§ 6. This is immediately evident if we keep in mind the fact
that algebraic expressions are symbols for numbers and that if the
letters are replaced by numbers, the algebraic fraction becomes a
numerical fraction.
43. Peinoiple I. Both numerator and denonhinator of a frac-
tion may he multiplied (or divided) hy the same expression with-
out changing the value of the fraction.
This follows from (5), § 6.
44. Pkinciplb II. If the signs of loth numerator and denomi-
nator of a fraction he changed, the sign of the fraction remains
unchanged.
This follows from Principle I, when we multiply both numerator and
denominator by — 1 .
45. Peinciple III. If the sign of either numerator or denomi-
nator (but not both) he changed, the sign of the fraction is changed.
This follows from (6), § 6.
46. Reduction. A fraction is said to be reduced to its lowest
terms when its numerator and denominator have no common
factor. We effect this reduction by the following
EuLE. Divide hoth numerator and denominator hy their
highest common factor.
27
28 ALGEBRA TO QUADRATICS
EXERCISES
Reduce the following to their lowest terms.
'■4aa;2-8a6a; + 4ai2"
12ax2-12a62 12a(a: - 6)(a + 6)
Solution
4as2- 8a6x + 4a62 4a(x-6)2
_3(a; + 6)
H.C.F. =4a(x-6).
aS _ fts „ x8 - 2 a;2 + 2 X . lOa; + 2 ax ■
3.
6.
9.
9
X —
a;8 _ 2 x2 + 2 X
x^ + 4x
6x2 — 8 ax +2 a^!
X2-o2
a2 + 62 - c2 + 2 a6
a2 - 62 + c" + 2 ac
X* — xs — X + 1
a — 2ax — lOx + 5
_ (g - 5)2 6x2-8ax+2a2 _ 2 a26 + 2 a62 - 2 a5e
■ a2-62' ■ x2_a2 ' ■ 36c2-362c-3a6c"
g xi2 - a" g a2 + 62 - c2 + 2 a6 - . 21x° -9x2 +7x -3
x" - ai8 ■ ■ a2-62 + c2 + 2ac' ' 3x»+ 15x2 + x + 6*
-. 2x2+ 3a;- 9 x« - x8 - X + 1 ,_ x» + Sox' +3a2x + a'
x2-9 ' ■ 2x*-x8-2x + l' ■ a2 + 2 dx + x2
47. Least common denominator of several fractions. We have
the following
EuLE. Find the least common multiple of the various denomi-
nators.
Multiply loth numerator and denominator of each fraction hy
the expression which will make the new denominator the least
common multiple of the denominators.
EXERCISES
Reduce the following to their least common denominator.
,2 3 ^ 2x-3
1. -) , and
X 2x-l 4x2-1
Solution : The L.C.M. of the denominators is x (4x2 — 1). Thus the frac-
tions are
2(4x2-1) 3x(2x + l) ^ x(2x-3)
1 and
x(4x2-l) x(4x2-l) x(4x2-l).
i- ) r 1 and — — —. i. , 1 , and
6-a6 + a a2-62 2x-3 4x2 + 4x-15 4x2 — 26
FRACTIONS 29
6.
1
1
and —
1
x'
-y-
xi
-
t'
a
^
6
w
and —
(a
-a2
02
-62'
2a!-
1
X
— , and
2s-
-3
a;2-2a; + l x^-1 (x + 1)"
7. ; , ; , and
a + b-c + 6 + C a.^ + 2ab + b^-c^
48. Addition of fractions. This operation we perform as
follows :
EuLE. Reduce the fraetions to he added to their least common
denominator.
Add the numerators for the numerator of the sum, and take
the least common denominator for its denominator.
49. Subtraction of fractions. This operation we perform as
follows :
Etile. Beduce the fractions to their least common denomi-
nator.
Subtract the numerator of the subtrahend from that of the
minuend for the nv/merator of the result, and take the least
common denominator . for its denominator.
50. Multiplication of fractions. This operation we perform
as follows :
EuLE. Multiply the numerators together for the numerator
of the product, and the denominators for its denominator.
51. Division of fractions. This operation we perform as
follows :
Etjle. Invert the terms of the divisor and multiply by the
dividend. o
ct c b
Bemabe. Since a fraction is a means of indicating division, 7 "^ "; and —
are two expressions for the same thing. -
a
30 ALGEBRA TO QUADRATICS
EXERCISES
Perform the indicated operations and bring the results into their simplest
forms.
0+6 a— 6
a — b g + 6
a+b a—b
a—b a+b
Solution :
a + b a-b (a + b)^ + (a- b)^
a-b a + b a? -Ifi
a + b a-b (a + b)^ - (a - 5)^
a-b a + b a^-b»
(g + 6)g + (g - 6)" a" - 6"
"" g2 - &2 ' (g + 6)2 - (g - 6)2
_ 2a^ + 2b^ _ a^ + b^
" 4 g5 2 g6
1-1 3 (¥)' - 1 4 f + A
i + i' ' 2-¥ ' ■ i-'i-A
2--V- g 3 + ^ + i y 2 + » + l
!-(¥)'■ ■ 1+1 + A ■ 1-f + l
^
^a_a + c 9. ^_1._1. 10. JL+.l
6 25 'g6ac6c g — 6g + 6
11.1+1+1. 12. »-Ill^+L^ 13. 2== ^
g5c ' c 'g-lga-l
,. Se-I 2a-7 IK 2a;-l 2a;-5
14. ■ la.
1 — Zx 7 a;-2x — 4
16.-5 ^. 17. ^ «
4x-4 6a; + 6 ' 3x-9 5a;-15
18 « + & g' + 2a6-62 -_ 2g-36 3g-26
■ g-6 ■ g2-62 ■ '120 15g
oQ 8 9 „- at? + 6c gd — 6c
15(a;-l) 10(a + l) ' 2cd(c-(J) 2c(J(c + d)
22 ''' + y ^-y ^'"y 23 ^" + 3' ('' + &) + aft g' — g''
■ a;-2/ K + y 3?-y^ ' x"^ - x{a + b) + ab nfi - V^'
24. i5 + -il + H£-l^. 25 5 8
6a 14a 35g 15a ' g^ - 9g + 14 ' g^ - 5a-14'
26. f^ + lU/^-l + lV 9.1 °(°-») g(g + g)
\y» xl V 2/ */ ' g'' + 2ga! + x2 a2-2ax + a!i'
28. L:^.i^.(i + -^). 29. ^_ + _J_ + _^_l.
1 + 5 o + o« \ l-g/ (x-l)«^(x-l)2^a;-l »
FRACTIONS 31
30. «+-!-■ 31. 1^*Z 32 "
33. L- 34. 35. a + 5
x + 4 + -1- c + -i-
3g + 6 a-b a^-l^ g_ 1 + a: 1-g g g^-S"
11 ''l l''ll "aFW
(g + 6)2 (a - 6)2 1-x 1 + a; g 6
39 ^-^V ^ (C + 3y ^^ __1 x + 2
' x^-2xy-16i/' :ifl~8xy+16y^' ■3(a; + l) 3(- 4- 3a; + x2)'
X y + z
42 2g-36 + 4 3g-46 + 6 g-1
6 8 12
43 / 'g' + 2/^ a:^ - y2 \ / g 4, y "^ - ^ V
\a;2 — 2/2 x^ + yV ' \x-y x + y)
x-1 y -1 z-1
.. 3a;2/z a; 2/ z
45.
yz + za; — a;?/ 11 1 -
X y z
/ 2x + y 2y-x _ x^ \ ^ a" + j
\x + y x — y x^ — yVx^-'i
.„ g — 354a — 65a + 3c g^ — 6c 2g
6a 26 9c 2ac 5"
, „ bed cda
47. • ; 1-
(a - 6) (a - c) (o - d) (6 - c) (6 -d){b- a)
dab
(c — 6) (c — a) (c — d)
a6c
"*" (d - a) (d - 5) (d - c) '
48 — + ""^^ _ _?. _ 3a-4c2 , _^ , 5 c^ - S 5
■ 6a 3a6 45 8ac2 Sc^ 126c2
49.^-^ + .-^ + : '
x-1 x + 1 (x-iy (1 + 1)8 x^-l (a;2 - 1)2
CHAPTEE IV
EQUATIONS
52. Introduction. An equation is a statement of equality
bet-ween two expressions.
We assTime the following
Axiom. If equals le added to, subtracted from, multiplied by,
or divided by equals, the results are equal.
As always, we exclude division by zero. In dividing an equation by an alge*
braic expression one must always note for what values of the letters the divisor
vanishes and exclude those values from the discussion.
53. Identities and equations of condition. Equations aie of
t-wo kinds :
First. Equations that may he reduced to the equation 1 = 1 by
performing the indicated operations are called identities.
Thus 2 = 2,
a — 5= (3a — 25) — (2a — 6)
are equations of this type. In identities the sign = is often replaced by =. It
should be noted that identities are true whatever numerical values the letters
may have.
Second. Equations that cannot he reduced to the form 1 = 1,
hut which are true only -when some of the letters have particular
values, are caUed equations of condition or simply equations.
Thus x = 2 cannot further be simplified, and is true only when x has the value
2. Also a; = 2 a is true only when x has the value 2 a or a has the value - ■ If in
2
this equation x is replaced by 2 a, the equation of condition reduces to an identity.
The number or expression -which on being substituted for a
letter in an equation reduces it to an identity is said to satisfy
the equation.
Thus the number 5 satisfies the equation i^ — 24 = 1. The number 3 satisfies
the equation (X — 3) (a; + 4) = 0.
32
EQUATIONS 33
The process of finding values that satisfy an equation is called
solving the equation. The development of methods for the solu-
tions of the various forms of equations is the most important
question that algebra considers.
In an equation in which there are two letters it may be possible
to find a value which substituted for either will satisfy the equar
tion. Thus the equation a; — 2 a = is satisfied if a; is replaced by
2 a, or if a is replaced by ^r- In the former case we have solved
for X, that is, have found a value that substituted for x satisfies
the equation. In the latter case we have solved for a. In any
equation it is necessary to know which letter we seek to replace
by a value that will satisfy the equation, that is, with respect to
which letter we shall solve the equation.
The letter with respect to which we solve an equation is called
the variable.
Values which substituted for the variable satisfy the equation
are called roots or solutions of the equation.
When only one letter, i.e. the variable, occurs in an equation, the root is a num-
ber. When letters other than the variable occur, the root is expressed in terms
of those letters.
54. Linear equations in one variable. An equation in which
the variable occurs only to the first degree is called a linear equa-
tion. To solve a linear equation in one variable we apply the
following
EuLE. Afply the axiom (§ 52) to obtain an equation in which
the variable is alone on the left-hand side of the equation.
The right-hand side is the desired solution.
To test the accuracy of the work substitute the solution in the
original equation and reduce to the identity 1=1.
Since the result of adding two numbers is a definite number, and the same is
true for the other operations used in finding the solution of a linear equation, it
appears that every linear equation in one variable has one and only one root.
When both sides of an equation have a common denominator, the numerators
are equal to each othfer. This appears from multiplying both sides of the equation
by the common denominator and then canceling it from both fractions.
34 ALGEBRA TO QUADRATICS
EXERCISES
Solve :
- ix — 2 , 5x 3x , _
Solution : Transpose the tenn involving x,
4a; — 2 6x 3x _
= 5.
5 8 4
• oji. x- 32x-16 + 25a;-30a! ,
Add fractions, = 5.
' 40
Clear of fractions and simplify, 27 s = 216.
2. ^(^±^ = ac+^.
dx d
Solution : Divide by a, __ — = ^ + 3-
ax a
Transpose the term involving x,
d2 + k2 X
■ — = c.
dx d
d2 + x2 - a;2
Add fractions,
dx
Clear of fractions and simplify, d = cx
d '
» = -.
c
3. (a-l)a; = 6-a;. 4. (a-x){l-x) = x^-l.
5. a(K-a«) = 6(a;-6=). 6. 2a! - |a; = f a; - J - fa; + 2.
7. 8s-7+a; = 9x-3-4s. 8. .617 a; - .617 = 12.34 -1.234 a;.
9. 3(2x- .3) = .6 + 5(a;-.l). 10. 7- 5x + 10 + 8a; - 7 + 3a; = a;.
11. (s-3)(a;-4) = (x-6)(a;-2). 12. f {T\K(|a;+5)-10]+3}-8=0.
13. (l + 6x)2 + (2 + 8x)2 = (l + 10x)2. 14. 5 = 3a;+i(x + 3)-Hllx-37).
15. 2(x + 5) (X + 2) = (2x + 7)(x 4- 3).
16. mx - 2 J) - m -h(^- 5x)] = is^.
17. 6x-7(ll-x) + ll = 4x-3(20-x).
18. (a - 6) (X - c) + (a + 6) (x + c) = 2 (&x + ad).
19. 2x - 3(5 + f x) + 1(4 - x) - \{Zx - 16) = 0.
20. 5x-2 = |x + |x + ^x + ^x + iJx + i-|».
21. (a - 6) (a - c + x) + (a + 6) (a + c - x) = 2 a«.
22. 12.9X - 1.45X - 3.29 - .99x - llx + .32 = 0.
23. 5.7x - 2^(7.8 - 9.3x) = 5.38 - 4i(.28 + 3.6x).
EQUATIONS 36
24. 3-| = -i-. 25. ^ + A = c.
3 11 mx nx
I x
26 ?jzi = 5jli 27 iii£^li)=?.
■ a:-3 K-5' ' J(5a; + 1) s'
28 li?^lil.=l. gg 2a:'-3g + S ^2
■f(3a;+6) 6' ■7a;2-4a;-2 7'
30.? + ^ = 12 + l 31. 1^%^ = ^-^.
cc 2 a; 9 i + x 4 J + a; -4
„„ a + 6a; _ c + da; 33 ^ 6x — 2 _ 5 2a; — 5
a + 6 "" c + d' ■ 3'7a;-3^ 7'3a;-7'
„. aa;ca;/ar^ „_a; + a6a; — 6,0
34. — + — + — = A. 35.—- = 1---
b d g b a a
36. 5^li^ + 6 = X - 1.
a
38.
3_1 2_1
2 a; 3 a;
2 X 3 a; 3 a;
40. ^^ + ^+l? = 2 + .^«
x+2 x-8 x+2
37.
3x-19 5x-25_
x-13 ' x+7 ~ ■
39.
a—x^b—x^c—x „
a 6 c
41.
0(2x4-1) 5ax-46_4
36 56 5
. „ ox 5x 2 o5 _ (o + 6)2x
'6 a + 6 06
43. 8Jx---3|a;-4ix + l = 0.
6
,. 1 a + 6 1 ,a — 6
44. + = ■ +
a+5 X a—b x
45 2(x-l) x + 8 _ 3(5x4-16)
■ x_7 x-4^ 5X-28'
, „ 0^6 — X , 6% — X , oc2 — X .
46. 1 1 = 0.
._ 5x-.4 1.3-3X l.S-Sx
6x4-2 15x4-5 3x41
49.I^^-i(x + 3) + 6 = §iH±^.
36 ALGEBRA TO QUADRATICS
5a; -1 3a; + 2 a;2_30x + 2
50.
3(a; + l) 2(a;-l) 6a;2-6
-, 3x-2 , 7a;-3 a; + 100
al. H = 10.
X-+S x + 2 a;2 + 5a; + 6
gg 3b(x-a) . =6-6? . 6(4a + ca) _Q
5o 156 6 a
,„ 16a; -27 a; +3 6 + 3a; 4a; -7
3^. . = .
21 5 2 3
54 a{b-x) ^ b(c-x)
bx ex
6x-6 9-lOx 3x-4 3-4x
X \c 0/
55.
56.
10 14 5 7
4-2x 4 1.5a; 4a;2
3 6a; -3 x-.5 8(2a;-l)
57.3-„,„^ =„,' ., + ; '
6(2a;-5) 2(2x-e) 3(2x-5)
-_ x''+i — Sx"-! 3x»-i-x» x» - ,
58. = 3x"-'.
4x 4 2
__ ax — 6c 6a; — ac _ ex — 6'' ^ ~ "• i-i *
ab c^ be e a
-„ X — 2a , X — 26 , X — 2c 3x
OO. ; 1 ; + -
6+c— o a+c— 6 a+6— c a+6+c
-, 2x» + 7x''-i 7x'' — 44k'»-i 4x» + 27x"-i
Dl. 1- -
62.
9 5x-14 18
3x + 3 /x + 1
2
a(x-3) 5(x-3) a^(x-l) 6^(x-l) ^
6 a 62 a2
64.
ac (m + ji)2x jix _ c 3jix
wi(o — 6)6 m6 6 m(a — 6) 6
-. 4(13x-.6) 3(1.2-x) _ 9x + .2 5 + 7x
5 2 ~20 4'
66. (°'' + ^') (3; _ a) + ?lll^(x - 6) = 2a(2a + 6 - X).
__ OX — 6 ex —d (6)1 + dm)x + {bp + dq) _ a e
' mx—p nx — q (mx — p){nx — q) ~ m n'
EQUATIONS 37
55. Solution of problems. The essential step in solving a
problem by algebra is the expression of the conditions of the
problem by algebraic symbols. This is, in fact, nothing else than
a translation of the problem from the English language into the
language of algebra. The translation should be made as close as
possible, clause by clause in most cases. In general the result
sought should be represented by the variable, which for that
reason is often called the unknown quantity.
Example. What number is it wliose third part exceeds its fourth part by
sixteen ?
Solution ; ' ' What number is It " is translated by x. Thus we let a; represent
the number sought. " Whose third part " is translated by - ■ " Exceeds its
XX
fourth part " is translated by , i.e. the third part less the fourth part
leaves something. ' ' By sijcteen ' ' gives us the amount of the remainder. Thus
the translation of the problem into algebraic language is
Let X represent the number sought.
5-5 = 16.
3 i
This equation should be solved and checked by the methods already given.
PROBLEMS
1. What number is it whose third and fifth parts together make 88 ? i,^
2. What number increased by 3 times itself and 5 times itself gives 99? C
3. What is the number whose third, fourth, sixth, and eighth parts
together are 3 less than the number itself?
4. What number is it whose double is 7 more than its fourth part ?
5. In 10 years a young man will be 3J times as old as his brother is
now. The brother is 1\ years old. How old is the young man ?
6. A father who is 53 years old is 3 years more than 12^ times as old as
his son. How old is the son ?
7. K you can tell how many apples I have in my basket, you may have
4 more than \, or, what is the same thing, 4 less than \ of them. How many
have I?
8. If Mr. A received \ more salary than at present, he would receive
$2100. How much does he receive?
9. A boy spends \ of his money in one store and ^ of what remains in
another, and has 24 cents left. How much had he ?
38 ALGEBRA TO QUADRATICS
10. A man who is 3 months past his fifty-fifth birthday is 4 J times as old
as liis son. How old is the son ?
11. In a school are four classes. In the first is ^ of all the pupils ; in the
second, J ; in the third, ^ ; in the fourth, 37. How many pupils are in the
school ?
12. A merchant sold to successive customers ^, ^, and J of the original
length of a piece of cloth. He had left 2 yards less than half. How long
was the piece ?
13. How may one divide 77 into two parts of which one is 2J times as
great as the other ?
14. The sum of two numbers is 73 and their difference is 15. What are
the numbers ?
15. A father is 4J times as old as his son. Father and son together
are 27 years younger than the grandfather, who is 71 years old. How old are
father and son ?
16. The sum of two numbers is 999. If one divides the first by 9 and
the second by 6, the sum of these quotients is 138. "What are the numbers ?
17. The first of two numbers whose sum is a is 6 times the second.
What are the numbers ?
18. If the city of A had 14,400 more inhabitants, it would have 3 times
as many as the city of B. Both A and B have together 12,800 more than
the city of C, where there are 172,800 inhabitants. How many are in
A and B ?
19. Two men who are 25 miles apart walk toward each other at the
rates of 3^ and 4 miles an hour respectively. After how long do they meet ?
20. A courier leaves a town riding at the rate of 6 miles an hour. Seven
hours later a second courier follows him at the rate of 10 miles an hour.
How soon is the first overtaken ?
21. A can copy 14 sheets of manuscript a day. When he had been work-
ing 6 days, B began, copying 18 sheets daily. How many sheets had each
written when B had finished as many as A ?
22. The pendulum of a clock swings 387 times in 5 minutes, while that
of a second clock swings 341 times in 3 minutes. After how long will the
second have swung 1632 times more than the first?
23. The difference in the squares of two numbers is 221. Their sum is
17. What are the numbers ?
24. If a book had 236 more pages it would have as many over 400 pages
as it now lacks of that number. How many pages has the book ?
25. A man is now 63 years old and his son 21. When was the father
19 times as old as his son ?
EQUATIONS 39
26. If 7 oranges cost as much less than 50 cents as 13 do more than
60 cents, how much do they cost apiece ?
27. The numerator of a fraction is 6 less than the denominator. Dimin-
ish both numerator and denominator by 1 and the fraction equals J. Find
the fraction.
28. The sum of three numbers is 100. The first and second are respec-
tively 9 and 7 greater than the third. What are the numbers ?
29. Out of 19 people there were f as many children as women, and IJ
times as many men as women. How many were there of each ?
30. A boy has twice as many brothers as sisters. His sister has 5 times
as many brothers as sisters. How many sons and daughters were th-ere ?
31. A dealer has 5000 gallons of alcohol which is 85% pure. He
wishes to add water so that it will be 75% pure. How much water must
he add?
32. How much water must be added to 5 quarts of acid which is 10% full
strength to make the mixture 8 J% full strength ?
33. A merchant estimated that his supply of coffee would last 12 weeks.
He sold on the average 18 pounds a week more than he expected, and it
lasted him 10 weeks. How much did he have ?
34. At what time between 3 and 4 o'clock are the hands of a clock point-
ing in the same direction ?
35. At what time between 11 and 12 o'clock is the minute hand at right
angles to the hour hand ?
36. A merchant bought cloth for $2 a yard, which he was obliged to sell
for $1.75 a yard. Since the piece contained 3 yards more than he expected,
he lost only 2%. How many yards actually in the piece ?
37. A man has three casks. If he fills the second out of the first, the
latter is still f full. If he fills the third out of the second, the latter is stUl
^ full. The second and third together hold 100 quarts less than the first.
How much does each hold ?
38. A crew that can cover 4 miles in 20 minutes if the water is still, can
row a mile downstream in | the time that it can row the mile upstream.
How rapid is the stream ?
39. A cask is emptied by three taps, the first of which could empty it
in 20 minutes, the second in 30 minutes, the third in 35 minutes. How long
is required for all three to empty the cask ?
40. A can dig a trench in | the time that B can ; B can dig it in f the
time that C can ; and A and C can dig it in 8 days. How long is required
by all working together?
40 ALGEBRA TO QUADRATICS
56. Linear equations in two variables. A simple equation in
one variable has one and only one solution, as we have already
seen (p. 33). On the other hand, an equation of the first degree
in two variables has many solutions.
For example, 3x + 7y = l
is satisfied by innumerable pairs of numbers which may be sub-
stituted for X and j/. For, transposing the term in y, we get
l-7y
. = -3—,
from which it appears that when y has any particular numerical
Talue the equation becomes a linear equation in x alone, and
hence has a solution. Thus, when y = 1, x = — 2, and this pair
of values is a solution of the equation. Similarly, a; = — 9, y = 4
also satisfy the equation.
57. Solution of a pair of equations. If in solving the equation
just considered, the values of x and y that one may use are no
longer unrestricted in range, but must also satisfy a second Unear
equation, we get usually only a single pair of solutions. Thus
if we seek a solution, that is, a pair of values of x and y satisfying
3x + 7y = l,
such that also .
x + y = -l,
we find that the pair of values a; = — 2, y = 1 satisfy both equa-
tions. Any other solution of the first equation, as, for instance,
x=—9, y = 4:, does not obey the condition imposed by the second.
Two equations which are not reducible to the same form are
called independent.
Thus 6x — Sy — i =
and 3x — iy = 2
are not independent, since the first is readily reduced to the second hy transposing
and dividing by 2. They are, in fact, essentially the same equation. On the other
X — iy= 2
and 3x — iy = 2
are not reducible to the same form and are independent. Since dependent equa-
tions are identical except for the arrangement of terms and some constant factor,
all their solutions are common to each other.
EQUATIONS 41
This principle we may state as follows :
Two eauations
and o'as + Vy + c' =
are dependent when and only when
— — — — £.
a' b' d
Independent equations in more than one variable -which have
a common solution are called simultaneous equations.
Two pairs of simultaneous equations which are satisfied by the
same pair (or pairs) of values of x and y and only these are called
equivalent.
Thus |3»= + 7« = 1. ^^ ra==-2,
are equivalent pairs of equations.
58. Independent equations. We now prove the following
Theoeem. If A=0 and B = represent two independent
equations, then the pairs of equations
^ = ^'(1) and |«^ + f = ^. (2)
B = ^ ' \cA-\-dB = ^ '
are equivalent where a, h, c, and d are any numhers such that
ad — he is not equal to zero.
The letters A and B symbolize linear expressions in x and y.
Evidently any pair of values of x and y that makes both 4 = and
5 = 0, i.e. satisfies (1), also makes aA-\-hB= and oA-\-dB = 0,
i.e. also satisfies (2). We must also show that any values of x
and y that satisfy (2) also satisfy (1).
For a certain pair of values of x and y let
aA-{-lB = 0, (3)
oA-\-dB= 0. (4)
Multiply (3) by c and (4) by a (§ 52).
Then acA + loB = 0, (5)
aoA + adB = 0. (6)
42 ALGEBRA TO QUADRATICS
Subtract (6) from (6) (§ 52),
{ad—he)B = Q.
Thus, by § 5, either ad — ho = or B = 0.
But ad — be is not zero, by hypothesis ; consequently B = 0.
Similarly we could show that ^ s 0.
Thus if we seek the solution of a pair of equations A = Q,
B = 0, we may obtain by use of this theorem a pair of equiva-
lent equations whose solution is evident, and find immediately
the solution of the original equations.
59. Solution of a pair of simultaneous linear equations. The
foregoing theorem affords the following
EuLB. Multiply each of the equations Toy some n/umher such
that the coefficients of one of the variables in the resulting pair
of equations are identical.
Subtract one equation from the other and solve the resulting
simple equation in one variable.
Find the value of the other variable by substituting the value
just found in one of the original equations.
Gheck the result by substituting the values found for both
variables in the other equation.
Example.
Solve 3x + 7y = l, (1)
x + y=-l. (2)
Solution : Multiply (1) by 1 and (2) by 3,
Zx + 1y = \,
Zx + Zy = -
Subtract,
4y = 4
Substitute in (2),
y = l.
x + \=-X.
Check: Substitute
in
(1),
x=-2.
3.
(-
-2) + 7-l = -6-
+ 7 = 1.
60. Incompatible equations. Equations in more than one vari-
able that do not have any common solution are called incompatible.
EQUATIONS 43
Thioeem. The equations
ax+bi/ = c, (1)
a'x + h'y = c' (2)
are incompatihle when and only when ah' — ha' = 0.
Apply the rule of § 59 to find the solution of these equations.
Multiply (1) by a' and (2) by a.
We obtain aa'x + a'by = ca',
aa'x + ab'y = ac'.
Subtract, (ab' — a'b) y = ae' — ca'.
If now ab' — a'b is not zero, -we get a value of y ; but since under
our hypothesis ab' — a'b = 0, we can get no value for y since divi-
sion by zero is ruled out (§ 7). Thus no solution of (1) and (2)
exists.
Example.
Solve 3a; + 7sf = l, (1)
&x + \iy = \. (2)
Solution : Multiply (1) by 2,
6a; + 142^ = 2
6a: + liy= 1
Subtract, = 1
which is absurd. Thus no solution exists.
61. R6sum£. We observe that pairs of equations of the form
ax -\- by -\- c = 0,
a'x + b'y + c' = Q
fall into three classes :
(a) Dependent equations, which have innumerable common
solutions. ^ J g
Tlien a' = b'^7'- W
(b) Incompatible equations, which have no common solution.
Then ^j, _ ^,5 ^ q^ ^^^. ^j-j -^ ^^^ ^^.^^
(c) Simultaneous equations, which have one and only one pair
of solutions.
Then ab' — a'b ^ 0.
44 ALGEBRA TO QOADHATICS
EXERCISES
Solve and check the following :
, 2x + 5y = l, ^ 4.x-6y = 8, 3 6x + Sy = lS,
6x + 7y = 3. '^- fx-2/ = 4. 'x + iy = S.
. 7a -32/ = 27, , 2x-|^ = 4, ^ iy = ^x-l,
6x-6y = 0. 3x-ly = 0. ' ^^2/ = |x-l.
_ 5x-4y + l = 0, „ 3x + 4y = 253, g 5x + 3^ + 2 = 0,'
1.7x-2.2j/+7.9=a j/ = 5x. ' 3x + 2y + l = 0.
,„ x + my = o, jj x + y=|(5o + 6),
x-n2/ = 6. ' x-y= H<» + S&)-
,„ 2x-32/ = -5a, ,, f x - ^(2/ + 1) = 1,
^^- Sx-2y = -bb. i(x + l) + i(y-l) = 9.
3.5x + 2^2/=13 + 4^x-3.5y, ,- 3x + 2 2/ = Sa^ + a6 + 56',
^*- 2iyX + .8^ = 22i + .7x-3j2/. ' Sy + 2x = 6a^ - ab + bb^.
1 , 1_8
- + - — T)
X y 6
? + 5 = 3,
X 2^ ' X 2/ 6
Hint. Retain fractions.
X — c _a 3x + 1 _ 4
18. ^73^~6' 19- 4-2y~S'
x-y = a-h. x + y = l.
21." ''
^ + ^ = cx.
x+1 a+b+c
20.
' " ' = 2,
2X-2/+1
3x-2/ + l_5
x-y + 3
5 7
22.
x + 23/ 2x + 2/'
7 5
3x-2 6-2/
.9x-.72/+7.3_ ,
24.
13X-152/ + 17 " '
1.2x-.22/ + 8.9_
26.
13X-152/ + 17
X 1 y _ 1
a + 6 a — 6 a — 6
X J/ _. 1
23.^ + 1
X •
y-1
x + 2/ =
25.
X — 2/ =
a — 6 + c'
a + b — c
a — b^e
2(a2 + &2)
0^-62
4a&
aS-62
.^ 1^ = 6-
0+6 a—b a+6
g,^ a- 6 a-c
■ a^ - X , a2 - 2/
1 ^ = 6 + C.
32.
34.
EQUATIONS 45
y = i-3x + x^. (ix-1)(x-S) = y.
3v^ — 4-v^ = l. X + y = 2 Va.
a!V^ + j/V3 = 3V3, gg 4 VxTT - 5 VjT^ = 7,
xV3-2/V2 = 2V2. ' 3 Vx + 7 - 7 Vv - 7 = 2.
Vx Vy Vx-B Vy + S
<o-Ma; + w-^^±A±l x + l y + 2_ 2(a;-y)
(a-b)lx + (a+b)y] = a-b+l. S(x~S.)-4:(y -3) = 12{2y -x).
62. Solutions of problems involving two unknowns. Tlie same
principle of translation of the problem into algebraic symbols
should be followed here as in the solution of problems leading to
simple equations (p. 37).
PROBLEMS
1. The difierence between two numbers is 3^. Their sum is 9J. What
are the numbers ?
2. What are the numbers whose sum is a and whose difference is 6 ?
3. A man bought a pig and a cow for $100. If he had given $10 more for
the pig and $20 less for the cow, they would have cost him equal amounts.
What did he pay for each ?
4. Two baskets contain apples. There are 51 more in the first basket than
in the second. But if there were 3 times as many in the first and 7 times as
many in the second, there would be only 5 more in the first than in the
second. How many apples are there in each basket?
5. A says to B, "Give me $49 and we shall then have equal amounts."
B replied, "If you give me $49, I shall have 3 times as much as you.
How much had each?
6. A man had a silver and a gold watch and two chains, the value of the
chains being $9 and $25. The gold watch and the better chain are together
twice and a half as valuable as the silver watch and cheaper chain. The
gold watch and cheaper chain are- worth $2 more than the silver watch and
the better chain. What is the value of each watch ?
46 ALGEBRA TO QUADKATICS
7. What fraction is changed into J when both numerator and denomi-
nator are diminished by 7, and into its reciprocal when the numerator is
increased by 12 and the denominator decreased by 12 ?
8. A man bought 2 carriage horses and 5 work horses, paying in all
$1200. If he had paid $5 more for each work horse, a carriage horse
would have been only J more expensive than a work horse. How much
did each cost?
9. A man's money at interest yields him |540 yearly. If he had received
J% more interest, he would have had |60 more income. How much money
has he at interest ?
10. A man has two sums of money at interest, one at 4%, the other at 6%.
Together they yield $750. If both yielded 1% more interest, he would have
$166 more income. How large are the sums of money ?
11. A man has two sums of money at interest, the first at 4%, the second
at 3J%. The first yields as much in 21 months as the second does in 18
months. If he should receive \% less from the first and Ylo more from the
second, he would receive yearly |7 more interest from both sums. What
are the sums at interest?
12. What values have a mark and a ruble in our money if 38 rubles are
worth 14 cents less than 75 marks, and if a dollar and a ruble together
make 6^ marks ?
13. A chemist has two kinds of acid. He finds that 23 parts of one kind
mixed with 47 parts of the other give an acid of 84^% strength and that
43 parts of the first with 17 parts of the second give an 80f% pure mixture.
What per cent pure are the two acids ?
14. Two cities are 30 miles apart. If A leaves one city 2 hours earlier
than B leaves the other, they meet 2J hours after B starts. Had B started
2 hours earlier, they would have met 3 hours after he started. How many
miles per hour do they walk ?
15. The crown of Hiero of Syracuse, which was part gold and part silver,
weighed 20 pounds, and lost \\ pounds when weighed in water. How much
gold and how much silver did it contain if 19^ pounds of gold and lOJ
pounds of silver each lose one pound in water?
16. Two numbers which are written with the same two digits differ by
36. If we add to the lesser the sum of its tens digit and 4 times its units
digit, we obtain 100. What are the numbers ?
17. A company of 14 persons, men and women, spend $48. If each man
spends $4 and each woman $3, how many men and how many women are in
the company ?
EQUATIONS 47
63. Solution of linear equations in several variables. This
process is performed, as follows :
EuLE. Eliminate One variable from the equations taken in
pairs, thus giving a system of one less equation than at first
in one less variable.
Continue the process until the value of one variable is found.
The remaining variables may be found by substitution.
Special cases occur, as in the case ol two variables, where an infinite number
of solutions or no solutions exist. Where no solution exists one is led to a self-
contradictory equation on application of the rule. See exercise 17, p. 48.
EXERCISES
Solve and check the following :
a: + y + z = 9,
1. a; + 2!/ + 4z = 15,
Solution : a;+2/+«=9 a;+2/+iS=9
a: + 2y + 4z = 15 a; + 3y + 9z = 23
y + 3z= 6 22/ + 8z = 14
^ + 42= 7
j/ + 3z = 6
y + 4z = 7
z = l
2/ + 3 = 6.
2/ = 3.
a; + 3 + 1 = 9.
!C= 5.
Check : 5 + 9 + 9 = 23.
a; + y = 87, x,\y = !t,y,
2. a; + z=25, 3. 2a; + 2z=x2,
j/ + z = 22. 3z + 3^ = 22/.
Hint. Divide the equations by ly, xz, yz respectively.
x + y + z = n, x + y + z = 36,
i. x + z-y = lS, 5. 4:X = Sy,
z + z~2y = 1. 2k = 3z.
1.3a;- 1.9 v=.l, 2x+2y + z=a,
6. 1.7y-l.lz = .2, 7. 2y +2z + a; = 6,
2.9a;-2.1z = .3. 2z+2x + y = c.
48 ALGEBRA TO QUADRATICS
x + y = m,
y + z = a,
« + tt = n,
u — x = b.
x + 2y = 5,
8.
2/ + 2Z =8,
z + 2 u = 11,
w + 2a; =6.
l + l = 2a,
y 2
10.
1 + 1 = 25,
X «
1 + 1 = 2.
X y
^y _oo
iy-Zx-"^'
12.
*2 IK
2x-3z = ^^'
2'* -12.
xy 1
x + y 5
.. xz _\
' s + z~6'
yz _ 1
j^ + z~7'
y + 1 '
13.?^ = 4,
z + 1 '
z + 3_l
iy-bz ' x + 1 2
2is = y + z + 8, x + 2/ = liz + 8,
14. 3f J/ = X + z + 12, 15. X + z = 2|2/ - 14,
4iz = x + y + 15. 2/ + z = 3fx-32.
x + 2y-z =4^6, x + 2y + 5z=15,
16. y + 2z-x = 10.1, 17. 3x + 52/ + 7z = 37,
z + 2x-y = 5.7. 5x + 8y + llz = 69.
7x + 6y + 7z = 100, (x+2)(22/+l) = (2x+7)y,
18. x-22/ + z = 0, 19. (x-2)(3z + l) = (x+3)(3z-l),
3x + y - 2z = 0. (1 +l)(«+2) = (j^+3)(z+l).
CHAPTEE V
RATIO AND PROPORTION
64. Ratio. The ratio of one of two numbers to the other is the
result of dividing one of them by the other.
The ratio of a to 6 is denoted by a : 6 or by 7 •
The dividend in this implied division is called the antecedent,
the divisor is called the consequent.
65. Proportion. Eour numbers, a, h, c, d, are in proportion whSn
the ratio of the first pair equals the ratio of the second pair.
a c
This is denoted by a ; 6 = c : (J or by - = — ■
The letters a and d are called the extremes, b and the means,
of the proportion.
66. Theorems concerning proportion. If a, i, c, d are in pro-
portion, that is, if
a:b = o:doT:'^ = ^, (I)
then ad = be, (II)
b:a^d:c, (III)
a:e = b:d, (IV)
a -\-b •.a = -\- d:c, (V)
a — b la^ — d:c, (VI)
a + b:a — b = c-{-d:c — d. (VII)
Equation (III) is said to be derived from (I) by inversion.
Equation (IV) is' said to be derived from (I) by alternation.
Equation (V) is said to be derived from (I) by composition.
Equation (VI) is said to be derived from (I) by division.
Equation (VII) is. said to be derived from (I) by composition
and division.
49
50 ALGEBRA TO QUADRATICS
67. Theorem. If a mimher of ratios are equal, the sum of any
number of antecedents is to any antecedent as the sum of the
corresponding consequents is to the corresponding consequent.
Let a:b = G:d = e:f= g -.h,
or
a c e g
b~ d~ f~ h'
To prove
a+o+e b+d+f
g h
If
a c e g
b~ d~f~ h~^'
we have
a = br,
e = dr.
•
e=fr.
g = hr.
Divide the
sum of the first three equations by the last and we get
a+c+e b+d+f
g n,
68. Mean proportion. The mean proportional between two mirn-
bers a and c is the number b, such that
a : 6 = 6 : c.
By (11), § 66, we see that aa = 62.
EXERCISES
It a:b = c:d, prove that :
1. a + b: = c + d : -•
a+b c+d
Solution: By (V), §66, " + ^ ' + "^
Squaring, we get
a c
(o + 6)" ^ (e + di)g
a2 c2 '
a + b _ c + d
a+b c+d
a + b: = c + d:
a+b c+d
RATIO AND PROPORTION 51
2. a2:62 = c2:d2. 3.a + b:c + d = a:c.
4. ma : mb = nc : nd. 5. a^:c^ = a^ + b'':(^ + d".
6. flS + 62 : -^ = c' + d^ : 7. Va^ + c^ : V6M^ = a : 6.
a + c + d
8. ma + nb:ra + sb = mc + nd:rc + sd.
9. a + b + c + d: a — b + c — d = a + b — c — d: a — b — c + d.
10. rind the mean proportional between a^ + c^ and b^ + d^.
11. Mnd the mean proportional between a^ + b!' + c^ and V + c^ + d^.
Solve the following for x :
12. 20:95 = x:57.
14. x — ax:Va = Vx : x.
.„ Vx + 'I+Vx _ i+Vx
Vx+l —y/x i — Vx
Hint. Use composition and division.
18. }- db): erf) = (a + 6)2 : 3
\a—b / a+b
13.
8a5:a; = 6c: Ifoe.
15.
l-^^:l-3Vic = l;4.
17
a + 5a2-62_ a-5
a — b ab ac
CHAPTEE VI
IRRATIONAL NUMBERS AND RADICALS
69. Existence of irrational numbers. We have seen that in
order to solve any linear ec[uation or set of linear equations with
rational coefficients we need to make use only of the operations
of addition, subtraction, multiplication, and division. When,
however, we attempt to solve the equation of the second degree,
3^ = 2, we find that there is no rational number that satisfies it.
Assumption. A factor of one member of an identity between
integers is also a factor of the other member.
Thus let 2 . a = 5, where a and 6 are Integers. Then since 2 is a factor of the
left-hand member, it must also be contained in b.
Theorem. iVb rational number satisfies the equation a? = 2.
Suppose the rational number ^ he a fraction reduced to its
lowest terms which satisfies the equation. Then
\bj -^^
or a^ = 2 W. (1)
Thus, by the assumption, 2 is contained in a^, and hence in a.
Suppose <x = 2 a'.
Then by (1) 4 a'^ = 2 h^,
or 2 a'2 = W,
that is, 2 must also be contained in J, which contradicts the
hypothesis that 7 is a fraction reduced to its lowest terms.-
The fact that the equation a;^ = 2 has no rational solution is
analogous to the geometrical fact that the hypotenuse of an
isosceles right triangle is incommensurable with a leg.
52
IRRATIONAL NUMBKRS AND RADICALS 53
70. The practical necessity for irrational numbers. For tlie
practical purposes of the draughtsman, the surveyor, or the
machinist, the introduction of this irrational number is superflu-
ous, as no measuring rule can be made exact enough to distin-
guish between a length represented by a rational number and one
that cannot be so represented. As the draughtsman does not use
a mathematically perfect triangle, but one of rubber or -wood, it
is impossible to see ia the fact of geometrical incommensurability
just noted a practical demand from everyday life for the intro-
duction of the irrational number. In fact the irrational number
is a mathematical necessity, not a necessity for the laboratory or
draughtiag room, as are the fraction and the negative number.
We need irrational numbers because we cannot solve all quad-
ratic equations without them, and the practical utility of those
numbers comes only through the immense gain in mathematical
power which they bring.
71. Extraction of square root of polynomials. This process,
from which a method of extracting the square root of numbers is
immediately deduced, may be performed as follows :
EtTLE. Arrange the terms of the polynomial according to the
powers of some letter.
Extract the square root of the first term, write the result as
the first term of the root, and subtract its square from, the given
polynomAal.
Divide the first term of the remainder ly twice the root
already found, and add this quotient to the root and also to the
trial divisor, thus forming the complete divisor.
Multiply the complete divisor hy the last term of the root and
subtract the product from the last remainder.
If terms of the given polynomial still remain, find the luxt
term of the root hy dividing the first term of the remainder hy
twice the first term of the root, form the complete divisor, and
proceed as before until the desired number of terms of the root
have been fov/nd.
54 ALGEBRA TO QUADRATICS
EXERCISES
Extract the square root of the following :
1. a*-2ahi + 2aH^ -2ax' + xK
Solution: o* - 2 g^x + Sa^r" -2axO + x* \a'' - ax + !X^
of
2 a' — ax| - 2 a%; + 3 dhfl — 2 ax' + x*
- 2 g^x + g^"
2a'-2aa + xi'| 2 a'kc^ - 2 ox' + x*
2 ggx" - 2 ax° + X*
2. 1 + X. 3. 1 - X.
4. 3x2 -2x + x*- 2x8 + 1. • 5. x*-6x8 + 13x2-12x + 4.
6. X* + 3/* + 2x'y - 2xyS - xV- 7- 9«* - 12 x' + Six" - 20x + 25.
8. 49a*-42g86+37o2&2-12a6s+46*. 9. 2a6 - 2ac - 26c H- a^ + 62 + c'.
10. M*M)2 + D*u2 + 10*1)2 + 2 u'v^W + 2 d'kJ^u + 2 JoSm^D.
72. Extraction of square root of numbers. We haye the
following
EuLE. Separate the number into periods of two figures each,
beginning at the decimal point. Find the greatest number
whose square is contained in the left-hand period. This is the
first figure of the required root.
Subtract its square from the first period, and to the remainder
annex the next period of the number.
Divide this remainder, omitting the right-hand digit, by twice
the root already found, and annex the quotient to both root and
divisor, thus forming the complete divisor.
Multiply the complete divisor by the last digit of the root,
subtract the result from the dividend, and annex to the remainder
the next period for a new dividend.
Double the whole root now found for a new divisor and pro-
ceed as before until the desired number of digits in the root
have been found.
In applying this rule it often happens that the product of the complete divisor
and the last digit of the root is larger than the dividend. In such a case we must
diminish the last figure of the root by unity until we obtain a product which is
not greater than the dividend.
IRRATIONAL NUMBERS AND RADICALS
56
At any point in the process of extracting tlie square root of a numlier before
the exact sqaare root ia found, the square of the result already obtained is less
than the original number. If the last digit of the result be replaced by the next
higher one, the square of this number is greater than the original number.
Therearealwaystwovaluesof theequare rootof anynumber. Thus 'v4=+2
or — 2, since (+ 2)? = (— 2)2= 4. The positive root of any positive number or
expression is called the principal root. When no sign is written before the radical,
the principal root is assumed.
EXERCISES
Extract the square root of the following :
1. 2.0000.
Solution :
2'.00'00'00'|1.414
1
2.4|1.00
96
281|400
281
2.824|11900
11296
604
2. 95481.
3. 56169.
4.
8.
5. 877969.
6. 2949.5761.
7.
5.
8. 257049.
9. .00070128.
10.
99.
11. 69.8896.
12. .0009979281.
13.
12.
14. 49533444.
15. 9820.611801.
16.
160.
73. Approximation of irrational numbers. In the preceding
process of extracting the square root of 2 we never can obtain a
number wliose square is exactly 2, for we have seen that such a
number expressed as a rational (i.e. as a decimal) fraction does not
exist. But as we proceed we get a number whose square differs
less and less from 2.
Thus
1.2 = 1, less than 2 by 1.
1.42 = 1.96, less than 2 by .04.
1.41^ = 1.9881, less than 2 by .0119.
1.414" = 1.999396, less than 2 by .000604.
66 ALGEBRA TO QUADRATICS
Thougli we cannot say that 1.414 is the square root of 2, we may
say that 1.414 is the square root of 2 correct to three decimal
places, meaning that
(1.414)2 < 2 < (1.415)2.
74. Sequences. The exact value of the square root of most
numbers, as, for instance, 2, 3, 5, cannot be found exactly in deci-
mal form and so are usually expressed symbolically. By means
of the process of extracting square root, however, we can find a
number whose square is as near the given number as we may desire.
We may, in fact, assert that the succession or sequence of numbers
obtained by the process of extracting the square root of a number
defines the square root of that number. Thus the sequence of
numbers (1, 1.4, 1.41, 1.414, • • •) defines the square root of 2.
75. Operations on irrational numbers. Just as we defined the
laws of operation on the fraction and negative numbers (pp. 2-4),
we should now define the meaning of the sum, difference, prod-
uct, and quotient of the numbers defined by the sequence of num-
bers, obtained by the square-root process. To define and explain,
completely the operations on irrational numbers is beyond the
scope of this chapter. It turns out, however, that the number
defined by a sequence is the limiting value of the rational num-
bers that constitute that sequence, that is, it is a value from which
every number in the sequence beyond a certain point differs by as
little as we please. We may, however, make the following state-
ment regarding the multiplication of irrational numbers : In
the sequence defining the square root of 2, namely, (1, 1.4, 1.41,
1.414, • • •) we saw that we could obtain a number very nearly
equal to 2 by multiplying 1.414 by itself. In general, we multi'
ply numbers defined hy sequences by multiplying the elements of
these sequences; the new sequence, consisting of the products, defines
the product of the origiri,al numbers.
Thus (1, 1.4, 1.41, 1.414, ■ ■ ■) (1, 1.4, 1.41, 1.414, • ■ •)
= (1, 1.96, 1.9881, 1.999396, • • ■).
The numbers in this Sequence approach 2 as a limit, and hence
the sequence may be said to represent 2.
IRRATIONAL NUMBERS AND RADICALS 57
76. Notation. We denote the square root of a (-where a repre-
sents any number or expression) symbolically by Va, and assert
Va- sfa ={-sfaf = a,
or, more generally, _ _ .
V a • ■y/b = V a • b.
Similarly, Va -h Vs = Vas -i- h.
EXERCISES
1. Form five elements of a sequence defining VS.
2. Form five elements of a sequence defining V5.
3. Form five elements of a sequence defining V6.
4. Form, in accordance with the rule just given, four elements of the
sequence ■v'2 ■ V3. Compare the result vrith the elements obtained in Ex. 3.
5. Form similarly the first four elements of product V2 • \/5 with the
first four elements obtained by extracting the square root of 10.
77. Other irrational numbers. The cube root and higher roots
of numbers could also be found by processes analogous to the
method employed in finding the square root, but as they are
almost never used practically, they will not be included here.
It should be kept in mind, however, that by these processes
sequences of numbers may be derived that define the various
roots of numbers precisely as the sequences derived in the pre-
ceding paragraphs define the square root of numbers.
The rath root of any expression a is symbolized by Va. Here
n is sometimes called the index of the radicaL The principle for
the multiplication and division of radicals with any integral
index is given by the following
Assumption. The, 'product {or quotient) of the nth root of two
numbers is equal to the nth root of the product (or quotient) of
the numbers.
Symbolically expressed,
nr— Hi— ni
Wa- -vb = ■y/a-b,
Va -H Vs = Va -f- b.
68 ALGEBKA TO QUADRATICS
78. Reduction of a radical to its simplest form. A. radical is
in its simplest form when the expression under the sign is integral
(§ 11) and contains no factor raised to a power which equals
the index of the radical ; in other words, when no factor can be
removed from under the radical sign and still leave an integral
expression. We may reduce a quadratic radical to its simplest
form hy the following
Etjle. If the expression under the radical sign is fractional,
multiply both numerator and denominator by some expression
that will make the denominator a perfect square.
Factor the expression under the radical into two factors, one
of which is the greatest square factor that it contains.
Take the square root of the factor that is a perfect square, and
express the multiplication of the result by the remaining factor
under the radical sign.
If the radical is of the nth index, the denominator must be made a perfect nth
power, and any factor that is to be taken from under the radical sign must also be a
perfect nth power.
EXERCISES
Beduce to simplest form :
1. V¥-
o, ^ /i2 /12T5 /4TT5 2 ^
solution: ^_ = ^__ = ^__ = _ Vl5.
2. Vi- 3. V^. 4. Vf-
5. y/Wl. 6. VA- 7- v^33.
8. v'250. 9. Vr+4. 10. VF^-
11. 8V75. 12. i^f^. ■ 13. I VSO^.
14. VA+^- 15- V¥ + l- 16. J/VTT^.
/ 1 \k.^ foe 1.
17.
20.
23.
IRRATIONAL NUMBERS AND RADICALS 59
26. Vx»-2a;2y + a;3/2. 27. V5 a^ - 20 a;" + 20 a.
28. jtll^±£. 29. ^H
•\ tKC2^.65;2 \602^
30. . ^ZZIZ+I". 31. ^H
\8a;-8a;2 + 2a;8 \
-12!i:'' + 18g
2^-202/2 + 2 2^8'
9{a-b)
79. Addition and subtraction of radicals. Radicals that are of
the same index and have the same expression under the radical
sign are similar. Only similar radicals can be united into one
term by addition and subtraction. We add radical expressions by
the following
EuLE. Reduce the radicals to he added to their simplest form.
Add the coefficients of similar radicals and prefix this sum
as the coefficient of the corresponding radical in the result.
A rule precisely similar is followed in subtracting radical
expressions.
EXERCISES
Add the following :
1. V27, Vis, and VtB.
Solution : V27 = V 9-3 = 3 V3
Vi8 = VlF3= 4V3
■n/76 = v'2ST3 = 6 V3
Sum = 12 V3
2. V3 + 2V3. 3. 8\/7-3V7. 4. a-Zx-b-yi.
5. a + 2 Va + 3 v^ + 2 Vila - v'27a.
6. 3 V8 + 4 V32 - 5 VBO - 7 V72 + 6 V98.
7. 8Va + 5Vx-7Va + 4Va-6-v^-3v'a.
8. 7 VJs + 4 V9i + 3 VSSa - 5 V36i - 2 y/Wx.
9. Vo^^ + Viea - 166 + y/ca? - 6a;2 - V9(a - 6).
10. 4 Va% - 3 Vfc^ + 2 Vc^ + V¥z - 2 V(6 + d)^x.
11. 6 -\^ + 3 V2a - 5 V3k - 2 Vli + Vl2i - Vl8i.
60 ALGEBRA TO QUADRATICS
80. Multiplication and division of radicals. Eor these pro
cesses we have the following
EuLE. Follow the usual laws of operation (§ 10), using also
the assumption of § 77.
Beduce each term of the result to its simplest form.
The operations of this section are limited to the case where
the radicals are of the same index. Radicals of different indices
as Vs and V2 must first be reduced to the same index. See § 87.
EXERCISES
1. Multiply V2 - V3 by V2 - Vs.
Solution : V2 - VS
2-V6-Vl6 + V2i
= 2-V6-4 + 2-\^
= - 2 + V6.
2. Diyide "^ + ^ by v^ + 4^.
■Vxy
x + y
Solution • V^ _ (v^--v^ + V^)(vS + Vy) *
y/x+y/y \'xy(Vx + y/y)
_ V^ -s/xy v^
Vxy y/xy -Vxy
'Ix
'■ily
.'/*_l + .-'/2'
iy \x
Carry out the indicated operations and simplify :
3. V10.V5. 4. v^-vlS.
5. V28 . V7. 6. Vf - VI-
7. (a-hV^''. 8. vii- V5!.
9. (V7-V3)(V3-V2). 10. (-I + VS)".
11. (5V3 + V6)(5>^-2). 12. Vli^.VTO^.
* Since a' + i' = (a + i)(a? - ab + b'), if a= Vx, b= -v^TrehaTO
IRRATIONAL NUMBERS AND RADICALS 61
13. (a + 6-Va5)(VS + Vb). 14. (8 + 3 Vs) (2 - VS).
15. V-v^ + Vy ■ Vvx-Vy. 16. V6 + 2V5. Ve - 2 V5.
17. ■^''s + Va"-! • -^x - Vk^^. 18. (a;2 + y^) -i- (a; -v^ + 2/ ■^).
19- -^ vS. 20. (4Va-V3^)(Va + 2V3a).
V^ W a;/ V-^ v^/
29. (2V6-Vl2-V2i + V48)v^.
30. (3V8 + Vi8+V50-2V72)V2.
31. (5v^-4V32 + 3V50-3V54)V3.
32. (V9kT5 + 3v^)(V9¥T5-3-\^).
33. [(V7 4 V3 + Vl0)(V7 + V3-Vi0)]''.
34. (2 V30^-3V5+ 6V3)(V8+V3-V5).
35. (2V5 + V8-Vi2)(^V30-|V3 + V2).
-(^^+^/¥)(^^-^/?)•
37. Pind the value of J V2i - Vf + 2 Vs - V5- V3 4- V5 to three
decimal places.
81. Rationalization. The process of rendering the irrational
numerator (or denominator) of a fractional expression rational
without altering the value of the' Traction is called the rationaliza-
tion of the numerator (or denominator) of the fraction.
This is usually accomplished by multiplying both numerator
and denominator of the given fraction by a properly chosen
radical expression called the rationalizing factor.
-* >
62 ALGEBRA TO QUADRATICS
The principles in accordance with which this rationalizing
factor is selected are the following:
Peinciple I Since (a + 6)(a — &)= a' — b% the rationalizing
factor of ■\fx ± ^Jy is Vaj T Vy.
Principle IL Since (a" — ab + 'b'^(a + b)= a' + b", the ration-
alizing factor of V* + ■^ is V^ — V^ + V^, and conversely.
Since (a^ + ab + b^){a — b)=a^ — &', the rationalizing factor
of -^ — -y/y is V^ + Vxy + -y/y^, and conversely.
EXERCISES
Eationalize the denominators of the following :
I Va +v^
Va —Vx
Solution : By Principle I the factor which will render the denominator
rational is Va + Vx.
Va + Vx Va + Vx Va + Vx a + x + 2Vca>
Thus
Va—Vx Va — Vx Va + Vx
1
2+V2+V3
Solution : This problem requires a twofold rationalization.
1 ^ , (2+V2)-V3
2 + V2 + V3 ~ [(2+ V2) + V3] [(2 + V2) - V3]
_ 2 + \^-VS _ 2 + V2-VS
~ (2 + V2y-S ~4 + 4V^ + 2-3
_ 2 + V2-\/3 _ (2H-V2--s/3)(3-4v^
~ 3 + 4V2 (3 + 4V2)(3-4-\^
_ 6 + 3\^-3V3-8V2-8 + 4V6
~ 9 - 16 . 2
2 + 5V2+3VS-iVB
8.
Solution :
^2+^6 + ^3 V4 + v'6 + V9
^-^ 'V2-^ ^ + ^ + 'V^ 2-3
IRRATIONAL NUMBERS AND RADICALS 63
7-V5 g V3 + V2
7.
2+V3 3 + V5 V3-V2
? 8 _? o 5Vi
10.
V3 + V2 V2-V4 V2 + 3VI
^£±^. H. ^^
\a-VS \a-
t + Vgg - 1
-Vo2-l
12 2V6 j3 I + 3V2-2V3
V5 + V3+V5 ' V6 + V3+V2
J. 2V15 „ Vd + a; + Va — a;
V3 + V5 + 2-v/2 Va + i-Va-x
jg 2 ^^ V6-V5_V3+V2
Va + 1 + Va - 1 V6 + V5-V3-V2
18. Show that "^ = - .10 ■ • ■.
V2 + V3
19. Show that ^^-^^ = 17.48 ... .
V8-
-V7
20
V(l + a)(H-6)-
-va
-a)(l
-6)
21.
V(l + a)(H-6) + V(l
Showthat^^ + ^-
-a)(l
V5-
-6)
V5_
.158.
Vs + V5 + Ve - V5
82. Solution of equations involving radicals. We prove the
important
Theokem. When an equation in x is multiplied ly an expres-
sion in X, the resulting equation has, in general, solutions uihich
■ the first one did not possess.
Let ^ =
represent an equation containing x which is satisfied by the
/values X = a, b, ■ ■ ■ n. Let £ be an expression which vanishes
when X = a, p, ■ ■ ■ V. Then the expression
A-B =
is satisfied not only when x = a, b, ■ ■ -, n, but also when
X = a, p, ■ ■ ; V.
64 ALGEBRA TO QUADRATICS
Example. The equation x — 2 =
has x = 2 for its only solution, while the equation
(X - 2) (s - 3) =
has in addition the solution a; = 3.
If in tlie course of a problem it is necessary to multiply an
equation by any expression inyolving the variable, the solutions
of the resulting equation must be substituted in the first one to
ascertain if any solutions have been introduced -which did not
satisfy the original equation. Solutions which have been intro-
duced in the process of solving an equation, but which do not
satisfy the original equation, are called extraneous solutions.
It may be shown in a similar way that raising the equation in *,
to any power introduces extraneous solutions.
EXERCISES
1. Solve Vx + 19 + Vs + lO = 9.
Solution : Vs + 19 = 9 - Vx + 10.
X + 19 = 81 + a; + 10 - 18 Va + 10.
- 72 = - 18Va! -H0.
4 = Va; + 10.
16 = a; + 10.
x = 6.
Check: V6 + 19 + V6 + 10 = 5 + 4 = 9.
2. Solve Vx + 19 - Vx + 10 = - 9.
Vx + 19 = - (9 - Vx.+ lO).
Simplifying, we get x = 6.
Check : VB + 19 - V6 + 10 = +5-4 = 1^-9.
Thus our result satisfies only the equation which was introduced in the
course of solving the problem, and is extraneous. The original equation
has no solution.
Solve and check, noting all extraneous solutions :
3. V3x -1=5. 4. VP -8 = 2.
5. V2x + V3x = l. 6. vx + V3x = 2.
7. Vsx-V = ^4x + 3. 8. 5v^-7=3\^-l.
IRRATIONAL NUMBERS AND RADICALS
65
9. Vx + 1 + Vx + 2 = 3.
11. 2^^-V^ = 2H-v^.
10. ■^/l3 + i^/x^^=6.
13. V37 - 7 Vbx^ = 4.
12. Va; + 4 + Va; + 1 = 1.
14. V'a;2-7a; + 19 = Va;-3.
15. 7 + Va;2 - 11 a; + 4 = x.
17. a; - Vaa; (1 + a;) + 1 - a; = 1.
19. |(7\^+5)-5=f(3Vi-l).
9 4
16. 8 + V(a; - 10) (a; - 5) = x.
18. -^2 (a; + 1) + V2a; + 15 = 13.
20. 2 V3 + 3-v^ = 3-\/2 + 2V3a
22.^+^ = 4.
24.
25-_3Va£
a + Vte 2 6 + 3 Vox
6x+ 10
26. V9x+10 =
V4X+9
l + VT^x 1-Vl-x
29. X — ax: y/x = -s/x:x.
28. Vo -x + V&-x=-
30. 2 V2xT2 + v^T^ =
12x4-4
V8x+ 8
31.
1 + 2V3X-5 11 + 2V3X-5
1 + 3V3X-5 11 + 5V3X-5
32. V9x + 7 + V4x + 1 = V25x + 14.
33. Vx + 15 + Vx - 24 - Vx - 13 = Vx.
34. Vx - 7 + Vx - 2 - Vx - 10 = Vx + 5.
35. (V5 - 7) ( V5 - 3) = (Vx - 6) (^^ - 5).
36. (a + Vx)Vx:(b--^)Vx = a + l:b-l.
37. (4v^-7):(5v^-6) = (v^-7):(^^-6).
38. ( V a Vft — V 6 Vajv^ = oV 6 Vx — b'way/i.
CHAPTER VII
THEORY OF INDICES
83. Negative exponents. We have already seen (§ 16) that
«»•«'» = »"+"' (1)
when n and m are positive integers. We now assume that this law
still holds when one or both of the numbers m and n are negative
or fractional.
If we let a— " = — >
a™
a" /I \
then — = «"( — I = «».«-"• = a"-"",
since the law (1) holds when n and m are any integers. This
notation may be expressed verbally as follows :
Principle. A factor of numerator or denominator of a frac-
tion may be changed from the numerator to the denominator,
or vice versa, if the sign of its exponent he changed.
84. Fractional exponents. Since (p. 57) Va • Va = a, it is
natural to devise a notation for va suggested by the law (1).
If we let Va = a^,
we have Va ■ Va = a^ ■ a^ = a^'^ ^ = a^ = a.
Furthermore, if we let Va = a"
it would be consistent with law (1) to write
1 11 11 t
[a"f = a''-a" = a'- » = a\
This notation we shall assume in general. Thus
1 m
66
THEORY OF INDICES 67
With the adoption of this notation we can attach a meaning
to any real number with any rational number for its exponent.
This notation may be expressed verbally in the following
Peestciple. The numerator of a fractional exponent indicates
a power, the denominator a root.
85. Further assumptions. The operation of multiplication is
subject to the following laws of exponents :
I. Commutative law of rational exponents :
(a^y = a"-'- = a"-" =(a'-y.
IL Associative law of rational exponents :
{a'^y = a'^-' = a'>-" = (a*)""'.
The laws of operation (§ 10) defined for integral values of the
symbols we also assume when the symbols are expressions with
rational exponents.
86. Theoeem. a'lf ={aby, where r is any rational number as -.
1
We raise both sides of the equation a'i'' = (aby to the jth
power separately and show that the results are equal.
Since r = -j
[iaby}" = liabfj = (ab)" = a^b".
p p p p pp p p
Also (arby = (am) = {a^^){a^b'') ■ ■ ■ (a^b'')
jt j^ V p if 1*
= (a«-a.«---a«)(5«-6'---*«)
» ^ 'V V — '
q terms g tenns
= {a^f{b^y = a^bK
Thus (ffl'-S'-)'' = [ W]'-
Extracting the gth root and taking the principal root, we obtain
a'^b'^ = (aby.
68 ALGEBRA TO QUADRATICS
EXERCISES
1. Express in simplest form with positive exponents :
36a-2&-ic-J
(a)
36a-26-»c-i
Solution : By Principle, § l_,
_ibc
~ a* ■
x'v-* v^ ■ Vi
V(a;lgy6)6 21g-lj/5iS-8
^ ' (a;i3/l)-J ' 35x-22/6«-»'
,: Ell /~~i — \ — ; „v 4s-"sf-s 15a-268-m
(3) xi2,i.J Vx-i2,-i.-. (k) ^-j^ . ^^^„^_, •
2. Arrange in order of magnitude tlie following : <* ^
(a) VI, ^, -^J^- ^ ^ ,
Solution : We first ask, Is (|)i > (f)* ? - f J:' N -> ] -^ ^
Raise both numbers to the sixth power. \.i^
We obtain (f)s and (f)2, ^i^ ^
or , f^ and f, "^i
er 2^^ and 2J.
Thus VI > Vf •
Now compare (|)* and (J)i.
Raise both numbers to the fourth power.
We obtain (f)2 and J,
or IJ and IJ.
Thus VI > VI.
THEORY OF INDICES 69
Now compare (|)^ and (|)1.
Raise both numbers to the twelfth power.
We obtain (|)* and ({)»,
or ^ and YjS
or 5y'j and 5||.
Thus -V^ > -V^.
The order oi magnitude is then Vli Vli Vf •
(b) V^, </i, </!■ (c) v^, V8, -t^.
(d) ■,J^, VI. (e) V3, V-S, ^.
3. Perform the indicated operations,
(a) V2 • V3.
Solution : V^ • V3 = 2^ ■ 3^ = 2^ • S'
= (22.38)* = V'4^=-^i08. ^ Y'-
V^-V6
(b) VJ . V5. (c)
^
Vs
,(d) ^^^ (e) ^^
Oi'
87. Operations with radical polynomials. These operations
follow the rules for the same operations previously given, pro-
vided the assumptions and principles of §§ 83-86 are observed.
EXERCISES
1. Divide x^ — y^ hy i/i — Vy.
2. Extract the square root of ix — 12xiyi + 9yi + S2xi — 48 yi + 64.
3. Simplify — — -— — -.
xi - Sxi + Zx-i-x-^
4. Divide -J^-V^ —JjV^hjsJ-.
a' Ifi
5. Extract the square root of 1 2.
6. Multiply - 3x-s + 2— by — - ^^.
a;* x' 6-'
7. Extract the square root of
4 „ 15
x^y-'-—yx-^ 4
49 2
8. Multiply Vk' - as + a' - — - + v^ -a; + v^-lbyv^+l.
— xl>y-'-—yx-^ + —yH-^ - —xy-^ + 26-.
QUADEATICS AND BEYOND
CHAPTEE VIII
QUADRATIC EQUATIONS
88. Definition. An equation that contains the second hut no
higher power of the variable is called a quadratic equation. The
most general form of the quadratic equation in one variable is
ax^ + bx + c = 0, (1)
where we shall always assume a, h, and c to represent rational
numbers, and where a^ 0. Every quadratic equation ia x can
be brought to this form by transposing and simplifying.
89. Solution of quadratic equations. The solution of a quad-
ratic equation consists in finding its roots, that is, the numbers
(or expressions involving the coef&cients in case the coefBcients
are literal) which satisfy the equation.
The common method of solving a quadratic equation consists in
bringing the member of the equa.tion that involves the variable
into the form of a perfect square, i.e. into the form
x^ + 2Ax + A''.
For example, let us solve
a!^ + 2a;-8 = 0.
Transpose 8, a;'' + 2 x = 8.
If now we add 1 to both sides of the equation, the left-hand
member will be a perfect square,
a'' -f 2 a: -f 1 = 9.
70
QUADRATIC EQUATIONS 71
Express as a square, (x + 1)* = 9.
Extract the square root, a; + 1 = ± 3.
Transpose, x = — 4 or 2.
Both — 4 and 2 satisfy the equation, as -we see on substituting
them for x. Thus
(- 4)'' + 2 (- 4) - 8 = 0,
and 2^ + 2 ■ 2 - 8 = 0.
Consider now the general case.
Let us solve ax"^ -\- ix -\- c = 0.
Transpose o, ax^ + bx =— c.
b c
Divide by a, x^ -\ — x =
a a
/by
Add I -7-- 1 to both members to make the left-hand member a
\2aJ
perfect square,
' b , b^ c ¥ -iao + b'
X^ + -X + -7-- = h -—Z = T— r
a 4 a-' a ia' ia''
/ , by b'-Aao
Express as a square, I x + ^r— I =
2aJ ia^
Extract the square root,
b _ -y/b^ — Aac
_, — 6 ± "s/P — iac * ,1^
Transpose, x = ( J-J
2 as
The roots are
- 6 + VJ2 - 4 ac -b-^b'-iac
x, = ) x, =
'■ 2a ^ 2a
That the equation can have no other roots appears from § 96.
* This expression for tlie roots, , , .j- — - —
X= : ,
2a
may be used as a formula for tlie solution of a quadratic equation.
Thus to solve the equation 2a:2-3a: — 6=0
we may substitute in the formula a= 2, 6 = — 3, c= — 6, and obtain
3±V9 + 48 3±VB7
x=
Thus
1 1
3+VB7 3-VS7
72 QUADRATICS AND BEYOND
One should verify the fact that both ^ ^^^
■' ,2a-
satisfy (1) and are consequently roots of the
equation. They are, in general, distinct from each other. For
particular values of the coefB^cients to he noted late r (§ 98) the
roots may be equal or complex (i.e. of form a + yS V— 1, where
a and /8 are ordinary rational or irrational numbers).
We may sum up the process of solving a quadratic equation in
the following
EuLE. Write the expression in the form asi? + 6a3 + c = 0.
Transpose the term Twt involving x to the right-hand side of
the equation.
Divide hath sides of tJ),e. equation by the coefficient of a?.
Add to both members the square of one half of the coefficient
of X, thus making the left-hand member a perfect square.
JRewrite the equation, expressing the, left-hand member as the
square of a binomial and the right-hand member in its simplest
form.
Extract the square root of both members of the equation, not
omitting the ± sign in the right-hand member.
Transpose the constant term, leaving x alone on the left-hand
side of the equation. The two values obtained on the right-hand
side by taking the + and — signs separately are the roots sought.
Check by substituting the solutionis in the original equation,
which should then reduce to an identity.
90. Pure quadratics. A quadratic equation in which the coeffi-
cient of the term in x is zero is often called a pure quadratic. Its
solution is found precisely as in the general case, excepting that
we do not need to complete the square. Thus let us solve
ax^ + c = 0.
Transpose c, ax^ = — c.
Divide by a, «' =
^ ' a
QUADRATIC EQUATIONS 73
Extract the square root, ^ =±\ •
\ a
The roots are x, =+\ > a;, = — \
\ a \ a
EXERCISES
Solve and check the following :
1. 3a;2-6x-10 = 0.
Solution : Transpose 10, 3x^ — 6x = 10.
Divide by 3, • x^-2x — ^.
Add the square of J the coefBcient of x, i.e. 1, to both sides,
x2 - 2 a; + 1 = Y- + 1 = ¥•
Express as a square, (x — 1)^ = J^.
Extract the square root, a; — 1 = ± V^-
X = 1 ± V^.
Check : 3(1 ±-^' - 6 (1 ± ^) - 10 = 0.
2. 8a;2 + 2E -3 = 0.
Solution : Transpose 3, fSx^ + 2x = S.
Divide by 8, a;^ + J a; = |.
Add the square of J the coefBcient of x, i.e. ^j, to both sides,
x' + ix + -,\:
Express as a square, * (x + ^Y = IJ-
Extract the square root, a; + J = ± f.
'I
or i
Check: 8 • (J)2 + 2.i-3 = i + l-3 = 0.
8 ( - 1)2 + 2 ( - I) - 3 = If - I - 3 = I - f - 3 = I - f = 0.
3. a;2 - aa; = 0. 4. x^ = 169.
^. x^-\x = \. 6. fa;2="560.
7. a;2 + a; - 1 = 0. 8. 19a;2 = 6491.
9. 3 a;2 - 7 a; = 16. 10. a;2 = .074529.
11. 3a;2+ ll = 5a;. 12. a;^ - IJ a; = 1.
13. a;2 + a; - 56 = 0. 14. 20a!2 + a; = 12.
74 QUADRATICS AND BEYOND
15. 7x2 + 9a. = 100. 16. 6x2 + 5x = 56.
17. 14x2 - 33 = 71x. 18. 5x2 + 13 = i4x.
19. x2 - 8x + 15 = 0. 20. 91x2 - 2x = 45.
21. x2 + 2 s - 68 = 0. 22. x2 - 6x + 16 = 0.
23. x2 - 10 X + 32 = 0. 24. 6 x2 + 26J = 25^x.
25. 6x2 - i3a;'+ 6 = 0. 26. 15x2 + 527 = 178 x.
27. 2 x2 + 15.9 = 13.6 X. 28. (x - 1)2 = a (x2 - 1).
29. a2(&-x)2=62^(j-x)2. 30. 13x2-19 = 7x2 + 6.
31. ax2 - (02 + l)x + a = 0. 32. (a - x)(x - 6) = - 06.
33. 14 x2 + 45. 5 X = - 36. 26. 34. 02 (a - a;)2 = 62 (6 - x)2.
35. (a-x)2+ (x-6)2 = a2 + 62. 36. (a-x)(x-6) = (a-x)(c-x>
ax2 c
37.
b d
„_ 15 X 810
«>y. — = — •
2 3x
.- 2x 1050
41. — =
3 7x
43.
45.
47.
49.
X + ll_2x + l
x + 3 " x + 5 '
ax + b _ mx — n
bx + a nx — m
0x2 — 6x + c _ c
Zx2 — mx + n n
X - 2 _ 3 (8 - x)
38.
2x + - = 3.
X
40.
x2 + | = 50.
42.
a + X 6 + X 5
6 + x a + x'~2'
44.
a;8_10a;2+l_
x2-6x+9 ~
46.
21^ 10 4 „
X x-2 x-3
48.
5x-7 14
9 +2X-3-" ^•
i;n
(a - x)8 + (X - 6)3 a' - 6'
3X+14
28-
- X
2x-3'^
5
x-1
= 12.
5 + x 8
-3x
2x
3-x
X
x-2
/a -XV
-«/«■
-^\ ^
51. — '— + — ^ = 12. 52.
2x- 3 x-1
_„ 5 + x 8 — 3x 2x -J
53. = 04.
(a — X) -
16 -X
4
2x-
-1
X —
2
5x-
-1
-(X-
-6)
a + b
2(x
-11)
x-4
X
-6
12
3x-
+ 1
5X-14
x-3
56. — - — + — - — = - + X - 1
9 5 X
57. a2 - x2 = (a - X) (6 + c - x).
58. ^ + I = ^.
1+vT^^ i-vr^ 9
,_2x + 2 12 X — 4 x-2
o". — — 1 7 = — \-
18 x + 4 4 6
60. (X - a + 6) (X - a + c) = (a - 6)2 - x2.
QUADRATIC EQUATIONS- 75
91. Solution of quadratic equations by factoring. When the
left-hand member of an equation can be factored readily, this is
the most convenient method of solution. It also illustrates very
clearly the meaning and property of the roots of the equation.
Example. Solve x^ + 2 e — 15 = 0.
Paetor the left-hand memher, (re + 5) (a; — 3) = 0.
The object in solving an equation is to find numbers that substituted for
the variable satisfy the equation. But since zero multiplied by any number
is zero (p. 3), any value of x which causes one factor of an expression to
vanish makes the whole expression vanish. If in this case a; = 3, our equa-
tion in factored form becomes
(3 -I- 5) {3 - 3) = (3 + 5) ■ =
and is satisfied. If we let a; = - 5, the other factor becomes zero, and the
equation reduces to the identity
(5- 6) (5 - 8) = (5 - 3) = 0.
Thus the numbers 3 and — 5 are solutions of the equation.
92. Solution of an equation by factoring. We have imme-
diately the
EuLB. Transpose all the terms to the left-hand memher of the
equation.
Factor that memher into linear factors.
The values of the variahle that make the factors vanish are
roots of the equation.
EXERCISES
Solve and check the following :
1. 6a;2 + a; = 15. 2. 6 a;^ -|- 7 a; = 3.
3. 5a;2_a;_6 = 0. 4. 5a;2 - 17x -I- 6= 0.
5. 13a;2_38a; = 3. 6. 2a;2 - 5a; - 25 = 0.
7. a? - 40x -I- 111 = 0. 8. 13a;2 - 40a; + 3 = 0.
9. a;2-18a;-208 = 0. 10. 3a;2 - 26a; + 35 = 0.
11. a;2 - 3aa; - 4a« = 0. 12. (x - af - (a; - 6)2 = 0.
13. (a;2 - 1) -t- (X - 1)2 = 0. 14. (3x - 5)2 - (9a; -1- 1)2 = 0.
15. (2x-l)(a; + 2)-Ka;-l)(x-2). = -4.
16. (7x-l)(a;-l-3)-(4a;-3)(a;-l)=24.
76 QUADRATICS AKD BEYOND
17. (3x + l)(z+l)- (4a; + 3)(a! - 1) = - 2.
18. (a;-a)(4aa;-6) + (a;-6)(4aa;-6) = 0.
19. 3a;-4-v^^^=2(x + 2).
Solution : Transpose, x — 4 = 4 Vx — 7.
Square, a;2_8x + 16 = 16x-112.
Transpose, x^ - 24 x + 128 = 0.
Factor, (x - 16) (x - 8) = 0.
The roots are x = 16, x = 8.
Check : 3 ■ 16 - 4 Vl6 - 7 - 2 (16 + 2) = 48 - 12 - 32 - 4 = 0.
3 • 8 - 4 V8^^ - 2(8 + 2) = 24 - 4 - 20 = 0.
In the following examples, as always, the quadratic equation should be solved
by factoring when possible. Recourse to the longer but sure method of completing
the square is always available.
When an equation is cleared of fractions or squared in the process of bringing
it into quadratic form (1), §88, extraneous solutions may be introduced. The
results should be verified in every case and extraneous solutions rejected.
20. o + Va^ - x2 = X. 21. VxTl = x-l.
22. 2x-V2x-l = x + 2. 23. 1- 6x + V6(x + 4) = 0.
24. Vll-x + Vx-2 = 3. 25. V2x + l-2V2x + 3 = 1.
Hint. Square twice. „„ , — ; — =— ^
^ 26. Va(x - 6) + v'6(x - a) = x.
27. VxTS + V2X-3 = 6. 28. 2 V3 + x - 4 V3-x = V60.
29. Vl + ox - VI - ox = X. 30. V6x-l-V8-2x = v'x -1.
31. Vx + 7 - -s/IsT^ = 3. 32. X - 10 = f (X - 1)-V2x-1-
33. Va2 - X + V62 + X = a + 6. 34. V4 - x + V5 - x = V9 - 2 x.
35. (a2-62)(x2+l) = 2(a2 + 52)x. 36. x + 2aV2{a^ + t^)-x=3a!'+m
31. xV^^=2 + 2V^^+2 = V^^+8. 38. ^^±^ + ../^ = 2^^.
\6 + x \6 — X \6
»„ Va + X + Va — x._ o ^j, Vtt + -v^_ 2-v^ (x+a)"
Va + X - Va — X x Va— Vx Va + -v^ a(x— o)
Hint. Rationalize.
.^ Va — X + Vx — 6 _ /a — X .„ ■- Vb-y/a _ 1 Va— V6
/a — X + Vx — 6 la — x .„ /- .
, ; = -v/ r- 42. Vx +
/a - X - Vx - 6 \x-6
Va — X — Vx — 6 \x— 6 V6 Vx Va
QUADRATIC EQUATIONS 7T
.- 2 o - (1 + gg) a; _ 2 6 + (1 + 6") a:
l + d^-Scuc ~ l + b^ + 2bx'
44. 2VgT4-3V2ai-3= A •
■Vx + i
45. {a - xY + (5 - 01)2 = f (a - x) (6 - a;).
46. abx,^ -(a + b) (db + l)a; + (a6 + 1)2 = 0.
47. V2a;-2+V3a; + 7 = V2a;+lH-V3a;-8.
48. V{a + a;) (a; + 6) + V(a -x)(x — b) = 2 VoK.
49: 2 V2a + 6 + 2a; - V10a + 6-6a; = V10a + 96-6a;.
93. Quadratic form. Any equation is in quadratic form if it
may be written as a trinomial consisting of a constant term and
two terms involving the variable (or an expression which may be
considered as the variable), the exponent in one term being twice
that in the other. By the constant term is meant the term not
containing the variable.
Thus a; — 8 V ^ + 13 = 0, g -i + a;-' — 3 = , o^aj-Z" - (o + 6) a;-" + 6^ = 0,
a;2 — 2 a; — 3 — Vaj^ — 2a! — 3 + 17 = are all in quadratic form. In tlie last the
whole expression a:^ — 2 a; — 3 is taken as the Tariable.
It is usually convenient to replace by a single letter the lower
power of the variable or expression with respect to which the
equation is in quadratic form.
EXERCISES
Solve and check the following :
. 1. a; - 8 V£ + 15 = 0. (1)
Solution : Let V^ = y.
Then x = yK
Substituting, (1) becomes y^ — Sy + 15 = 0.
Factor, (y - 5) (y - 3) = 0.
The roots are 2/ = 5, i/ = 3.
Thus -v^ = 5, v^ = 3,
or a; = 25, a; = 9.
Check :26-8-5 + 15 = 0;9-8-3 + 15 = 0.
78 QUADRATICS AND BEYOND
2. z-i-5xri + i =
= 0.
Solution: Let
x-S = y.
Then
x-i = y^.
Substituting, (1)
becomes
y2-52/ + 4 = 0.
Factor,
(y-4)(2/-l) = 0.
The roots are
y = 4, y = l.
Since
^i-^= '
rf ^
! have
' _4, 1 -1.
Thus v^ =
1
■4'
x^ =
4' " = ±§'
(1)
and v^ = -; x^ = l; a; = ±Vi = ±l.
3. 2Va;2 -2a;-3 + a;2-2a; = 6.
Solution : Add — 3 to both members and rearrange terms,
a;2 - 2a; - 3 + 2 Va;2 - 2x - 3 - 3 = 0. (1)
Let Vx2 - 2 X - 3 = 2/.
Then x2-2x- 3 = 2/2.
Substituting, (1) becomes y'^ + iy — S = 0.
Factor, . (V + 3) (y - 1) = 0.
The roots are y = —3, y = 1.
Hence 2/2 = x2 -2x - 3 = 9,
or x2-2x + l = 13.
Extract the square root, x — 1 = ± Vl3.
The roots are x = 1 ± Vl3.
Also x2-2x-3 = l,
or x2 - 2 X + 1 = 5.
Extract the square root, x — 1 = ± V5.
The roots are x = 1± VB.
4. x3 - 1 = 0. 5. x^-8 = 0.
6. x6-l = 0. 7. x» + 8x^ = 9x.
8. x-6x-i = l.
Hint. Divide by Vic. This factor cor'
responds to the root x = 0.
9. 4x* + 5x^-1 = 0. 10. 10x*-21 = xa.
QUADRATIC EQUATIONS 79
11. 2x*-3a;3 + a; = 0. 12. 3-t^ + 6v^=4.
13. aai'p + bxp + c = 0. 14. 3a;« - Ta;^ - 6 = 0.
15. 4a;« - 14a;3 + 6 = 0. 16. x* - 13a;2 + 36 = 0.
17. K- 12\^ + 11 = 0. 18. a;8 + 4i^ = lejaiv^.
19. \/S3-2Vi + a; = 0. 20. 8a;-6 + 999z-8 = 125.
21.1^ + 6x^^ = 66. 22. 2(-v^-3)"-3 = V5.
23. (a;2 - 10) (x^ - 3) = 78. 24. (x^ - ly + (x^ + 1) = 2.
25. (\/x-l)^ + v^ = V^. 26. a;S + a;S = (2S + 2-8)a;V^.
27. (t^ - 3)(v^ - 4) = 12. 28. (X + a)* + (cc + b)i = (a - 6)*.
29. (2a;2-3)2-(a;2 + 4)2 = 7. 30. 2 a;^ + 3 Va;^ - a; + 1 = 2 a; + 3.
31. (2a; + 3)4 + (2a; + 3)-J = 6. 32. 8(8a; - 5)8 + 5(5 - 8a;)6 = 85.
33. a;*-4(a+6)a;2+16(a-6)2=0. 34. 4 a;2 + 12 a; VTTs = 27 (1 + a;).
35. (2a;2-3a;+l)2=22a;2-83x+l. 36. (a;^- 5a; + 7)!' -(a;-2)(x-3) = l.
37. a;2 + 5 = 8a; + 2 Va;2 - 8a; + 40. 38. 2 ^{x - 4)2 + 4{a; - 4)"* = 9.
39. a;2 - - = 2. 40. Ezii = a; - 8.
x^ 2 + x
1 62
41. a;-» + = = 2. 42. a = a;2 + 6 + — .
xVx a;"
43. -v^ - 8 = —J 44. 1 + - - - '""
vS-2 '2 2(l + vTTx)^
._ J , 41v^ 97 , 5 .„ 2a; , (a;-
45. a;* -I = h a;*. 46. + ^
X ^2 (a; -1)2 2x
2a!
'(a;-l)2' y 2x
, =4a;2
Va;2 + 2
47. (s - 1)2 + 4(a; - 1) - 5 = 0. Hint. Lety=^^^^; theiii=^^
48. (a;2 + 2)S + -— L==4a;2 + 8.
94. Problems solvable by quadratic equations. The principle
of translating the problem into algebraic symbols, explained- in
§ 65, should be observed here. The result should be verified in
every case. It may happen that the problem implies restrictions
that are not expressed in the equation to which the problem
leads. In this case some of the solutions of the equation may
not be consistent with the problem ; for instance, when the
80 QUADRATICS AND BEYOND
variable stands for a nmnber of men fractional solutions should
be rejected. If only sucb results are obtained, the problem is
self -contradictory. Often negative solutions should be rejected,
as when the result indicates a negative number of digits in a
number. Imaginary or complex (p. 72) results in general mean
that the conditions of the problem cannot be realized.
PROBLEMS
1. The product of J and J of a certain number is 500. What is the
number ?
2. There are two numbers one of which is less than 100 by as much as
the other exceeds it. Their product is 9831. What are the numbers ?
3. The sum of the square roots of two numbers is V74. One of the
numbers is less than 37 by as much as the other exceeds it. What are the
numbers ?
4. A man sells oranges for ^\ as many cents apiece as he has oranges.
He sells out for |3. How many had he ?
5. If the perimeter of a square is 100 feet, how long is its diagonal ?
6. A man sells goods and makes as much per cent as ^ the number of
dollars in the buying price. He made $246. What was the buying price ?
7. Two bodies A and B move on the sides of a right angle. A is now
123 feet from the vertex and is moving away from it at the rate of 239 feet
per second. B is 239 feet from the vertex and moves toward it at a rate of
123 feet per second. At what time (past or future) are they 850 feet apart ?
8. What is the number twice whose square exceeds itself by 190 ?
9. What numbers have a sum equal to 53 and a product equal to 612 ?
10. The sum of the squares of two numbers whose difference is 12 is
found to be 1130. What are the numbers ?
11. By what number must one increase each factor of 24 • 20 so that the
product shall be 640 greater ?
12. What numbers have a quotient 4 and £* product 900?
13. Two numbers are in the ratio of 4 : 5. Increase each by 15 and the
difference of their squares is 999. What are the numbers ?
14. If ^ is divided by a certain number, the same result is obtained as
if the number had been subtracted from 4 J. What is the number?
15. Separate 900 into two parts such that the sum of their reciprocal
values is the reciprocal of 221.
QUADRATIC EQUATIONS 81
16. The denominator of a fraction is greater ty 4 than the numerator.
Decrease the numerator by 3 and increase the denominator by the same,
and the resulting fraction is half as great as the original one. What is the
original fraction ?
17. The numerator and denominator of a fraction are together equal
to 100. Increase the numerator by 18 and decrease the denominator by 16,
and the fraction is doubled. What is the fraction ?
18. A number consists of two digits whose sum is 10. Reverse the order
of the digits and multiply the resulting number by the original one, and
the result is 2944. What is the number?
19. The sum of two numbers is 200. The square root of one increased by
the other is 44. What are the numbers ?
20. The difference of two numbers is 10, and the difference of their cubes
is 20530. What are the numbers ?
21. Around a rectangular flower bed which is 3 yards by 4 yards there
extends a border of turf which is everywhere of equal breadth and whose
area is ten times the area of the bed. How wide is it ?
22. Two bicyclists travel toward each other, starting at the same time
from places 51 miles apart. One goes at the rate of 9 miles an Lour. The
number of miles per hour gone by the other is greater by 5 than the number
of hours before they meet. How far does each travel before they meet ?
23. A printed page has 15 lines more than the average number of letters
per line. If the number of lines is increased by 15, the number of letters per
line must be decreased by 10 in order that the amount of matter on the two
pages may be the same. How many letters are there on the page ?
24. A merchant buys goods for a certain sum. The cost of handling them
was 5% of their cost price. He sells for $504, gaining as much per cent as
j'if the cost price was in dollars. What was the cost price ?
25. A man had $8000 at interest. He increased his capital by $100 at the
end of each year, apart from his interest. At the beginning of the third year
he had $8982.80. What per cent interest did his money draw f
26. Two men A and B can dig a trench in 20 days. It would take A 9
days longer to dig it alone than it would B. How long would it take B
alone ?
27. A cistern is emptied by two pipes in 6 hours. How long would it
take each pipe to do the work if the first can do it in 5 hours less time than-
the second ?
28. A party procures lunch at a restaurant for $15. If there had been
5 less in the party, each member would have paid 15 cents more without
affecting the amount of the entire bill. How many were in the party ?
82 QUADRATICS AND BEYOXD
29. A party pays $12 for accommodations. Had there been 4 more in the
party, and if each person had paid 25 cents less, the bill would have been
115. How many were in the party ?
30. A grocer sells his stock of butter for $15. If he had had 5 pounds less
in stock, he would have been obliged to charge 10 cents more a pound to
realize the same amount. How many pounds had he in slock ?
31. A man buys lemons for $2. If he had received for that money 50
more lemons, they would have cost him 2 cents less apiece. What was the
price of each lemon ?
32. It took a number of men as many days to dig a ditch as there were
men. If there had been 6 more men, the work would have been done in 8
days. How many men were there ?
95. Theorems regarding quadratic equations. In this and the
following sections we prove several theorems eoncernuig quad-
ratic equations. Similar theorems are later proved in general for
equations of higher degree.
Theorem. If a is w root of the equation
aa^ + hx + c^O, (1)
then X — a is a factor of its left-hand member, and conversely.
The fact that a is a root of the equation is equivalent to the
assertion of the truth of the identity
aa^ + ba + c = 0,
by definition of the root of an equation (§ 53).
Divide ax^ + bx + chy x — a as follows :
X — a\ ax^ -\-hx -\- c\ax + (^ + aoQ
(J -f- aa) X -\- c
(b + aa) X — aa^ — ba
aa^ -\-ba + c
Since the remainder vanishes by hypothesis, ax^ ■\-bx-\-o\s
exactly divisible by a; — a.
Conversely, we have already seen (p. 3) that if a; — ar is a
factor of an equation, or is a root, since replacing a; by a would
make that factor vanish.
QUADRATIC EQUATIONS 83
EXERCISES
Form equations of which the following are roots.
1. 2, 6.
Solution : Since 2 and 6 are roots, (a; — 2) and (a; — 6) are factors, and the
equation having these as factors is-
(i - 2) (ic - 6) = !C2 - 8 X + 12 = 0.
2. 1, - 1. 3. - 3, - 4.
4. V2, - V2. 5. - 2, - 6.
6. - 4 V2, V32. 7. V27, - 3 V3.
8. 2 + V3, 2 - V3. 9. a + V6, a - V6.
96. Theokem. a quadratic equation has only two roots.
Given the equation ax^ + 6a; + c = with the roots a and /3, to prove that
the equation has no other root, as 7, distinct from a and p.
Since a and ^ are roots of the equation, x — a and a; — /3 are factors.
Thus our equation may be written in the form
o(a;-a)(a;-/3) = 0. (1)
If now 7 is a root, it must satisfy (1), i.e.
a(7-a)(7-/3) = 0.
But in order that any product of numerical factors should vanish one of
the factors must vanish. Thus either a = 0, or 7 — a = 0, or 7-/3 = 0.
But, by hypothesis, 1 ^ a and 7 5* ^, so the last two factors cannot vanish.
Thus a = 0. This would, however, reduce our equation to a linear equa-
tion, which is contrary to our hypothesis that the equation is quadratic.
Thus the assumption that we have a third distinct root leads to a contradiction.
CoEOLLARY I. If a quadratic equation is satisfied by more than two distinct
values of the variable, then each of the coefficients vanishes identically.
The above proof shows that the coefficient of x^ must vanish. In the same
way it can be shown that 6 = c = 0.
CoBOLLARY II. If two quodratic expressions have the same value for more
than two values of the variable, then their coefficients are identical.
Let ax,^ + bx + c = aV + b'x + c'
for more than two values of x. Transpose, and we obtain
{a - a')x^ + (6 - V)x + c - c' = 0.
We have then a quadratic equation satisfied by more than two values
of X. Thus by Corollary I each of its coefficients must vanish. Thus a' = a,
6' = 6, c' = c.
84 QUADRATICS AND BEYOND
This theorem taken with § 95 is equivalent to the statement that a quad-
ratic equation can be factored in one and only one way. Thus if
ax^ + 6x + c = a(x — a)(x — p),
we cannot find other numbers y and S distinct from a and p such that
ax^ + bx + c = a(x — y) {x — S),
for then the equation would have roots distinct from a and /3.
97. Theorem. If the equation
x^ + bx + c = 0, (1)
wliere 6 and c are integers, has rational roots, those roots must be integers.
P
Por suppose - to be a rational fraction reduced to its lowest terms and
a root of (1).
Then ^ + ^ + c = 0,
g2 q
or p2 + bpq + cg2 = 0,
which gives p^ = — q{bp + cq).
Thus some factor of q must be contained in p (§ 69), which contradicts
p
the hypothesis that the fraction - is already reduced to its lowest terms.
98. Nature of the roots of a quadratic equation. The equation
ax^ + bx + G==0 (1)
has as roots (§ 89) x^ = ~ ^ "*" ^t' ~ ^ "''' ' (^)
(^)
These expressions afford an immediate arithmetic means of
determining the nature of the roots of the given equation when
a, b, and c have numerical values and a # 0. In fact an inspec-
tion of the value of 6^ — 4 ac is sufllcient to determine the nature
of the roots of (1).
I. When ¥ — 4acis negative, ike roots are imaginary (§89).
II. When If — ^ac = 0, the roots are real and equal. In this
b
case x^ = x^=-—.
III. When ¥ — 4ci'C is positive, the roots are real and distinct.
rv. When W — Jj-ac is positive and a perfect square, the roots
are real, distinct, and rational.
-b + Vp-
- 4 ac
2a
_ 6 _ VS" -
- 4 ac
QUADRATIC EQUATIONS 85
In case IV, if J" — 4 ac = A,
_ --6+jv/a. _ -6- Va
2 OS 2 a
The conyerses of these four cases are also true. For instance,
if the roots of (1) are imaginary, from (2) and (3) it is clear
that S" — 4 ac must he negative.
The expression A = 6^ — 4 ac is called the discriminant of the
equation ax^ + bx + c = 0.
EXERCISES
1. Determine the nature of the roots of the following equations without
solving.
(a) Sx^ -4a; -1 =0.
Solution : A = (- 4)2 — 4 ■ 3 • (- 1) = 16 + 12 = 28 and is then positive.
Thus by III the roots are real and distinct.
(b) 3a;2 - 7x + 6 = 0. (c) 6x^-x-l = 0.
(d) 3a;2 + 4a; + l = 0. {e) x' - ix + 1 = 0.
(f) 2a;2-6x-9 = 0. (g) 2a:2 - 4a; - 2 = 0.
(h) 4a;2 + 12a; + 9 = 0. (i) 2a;2 + 6a; - 4 = 0.
(i) 4a;2- 28a; + 49 = 0. (k) 4a;2 + 12a; + 5 = 0.
2. Determine real values of k so that the roots of the following equations
may he equal. Check the result.
(a) (2 + &)a;2 + 2fca; + l = 0.
Solution : Here 2 + fc = a, 2k = b, 1 = c.
Thus A = 62-4a(; = 4J;2-4.(fc + 2).l
= ik^ - ik - 8.
Since the roots of an equation are equal when and only when its dis-
criminant eqiials zero (§ 98, II), the required values of k make A = and
are the roots of
442-44; -8 = 0,
01 }fi-k-2 = 0.
Solve by factoring,
k^-k-2 = (k-2){k + l) = 0.
Thus the values of k are fc = 2, ft = — 1.
Check : Substituting in the original equation for & = 2, we get
4a;2 + 4a; + l = (2a; + l)2,
and for & = —1 we get a;2 — 2a; + 1 = (x — 1)^.
86 QUADRATICS AND BEYOND
(b) x^ + kx+16 = 0. (c) x^ + 2x + k^ = 0.
(d) z2-2fcc + 1 = 0. (e) 3fca;2-4a;-2 = 0.
(f) &i2 - 3a; + 4 = 0. {s) x" + 4kx + k^ + 1 = 0.
(h) k'x^ + 3a; - 2 = 0. (i) (J;2 + 3)x2 + feB - 4 = 0.
(j) S*x= + fca; - 1 = 0. (k) x^ + (3fc + l)a; + 1 = 0.
(1) a:2 + 3a; + fc _ 1 = 0. (m) a;2 + 9 fex + 6fc + J = 0.
(n) ik^x" + ikx - 125 = 0. (o) 2x2 - 4x - 2fc + 3 = o.
(p) (fc + l)x2 + kx + k + 2 = 0. (q) fca;2 + (4fc + l)x + 4fc - 3 = 0.
(r) 2(A;+l)x2 + 3fcx + J:-1 = 0. (s) (fc - l)x2 + 5to + 6fc + 4 = 0.
(t) (2fe + 3)x2-7fcx + ^^^^ — !- = 0. (u) (fc-l)x2+(2fc + l)x+i; + 3 = 0.
CHAPTEE IX
GRAPHICAL REPRESENTATION
99. Representation of points on a line. Let us select on the
indefinite straight line AB a, certain point as a point of refer-
ence. Let us also select a certain line, the length of which for
the purpose in hand shall represent unity. Let us further agree
that positive numbers shall be represented on AB by points to
the right of 0, 'whose distances from are measured by the given
+1 +2 +3 1
B Unity
numbers, and negative numbers similarly by points to the left.
Then there are certainly on AB points which represent such num-
bers as 2, — 3, ^, — -J-l, or, in fact, any rational numbers. Since
we can divide a line into any desired number of equal parts, we
are able to find by geometrical construction the point correspond-
ing to any rational number. Furthermore, by the principle that
the square of the hypotenuse of a right triangle equals the sum
of the squares of the other two sides, we can find the point
corresponding to any irrational nuniber expressed by square-root
signs over rational numbers. More complicated irrational num-
bers cannot, however, in general be constructed by means of ruler
and compasses, but we assume that to every real number there
corresponds a point on the line, and conversely, -we assert that to
every point on the line corresponds a real numher. This assump-
tion of a one-to-one correspondence between points and real num-
bers is the basis of the graphical representation of algebraic
equations.
This amounts to nothing more than the assertion that every real number,
rational or irrational, as, for instance, — 6, 2 -j- v 3, V3, tt, represents a certain
distance from on AB, and conversely, that whatever point on the line we may
select, the distance from to that point may be expressed by a real number. .
87
88
QUADRATICS AND BEYOND
■'Tc
X axis
100. Cartesian coordinates. We have seen tliat when the
single letter x takes on real values all these values may be repre-
sented by points on a straight line.
"When, ho-wever, we have two variables,
as X and y, which we wish to represent
simultaneously, we make use of the
plane. Just as we determined arbitrar
rily, on the line along which the single
variable was represented, an arbitrary
point for the point of reference and an
arbitrary length for the unit distance,
so now we select an indefinite line along which x shall be repre-
sented, and another perpendicular to it along which y shall be
represented. The former we call the X axis ; the latter the Y axis.
The intersection of these axes we take as the point of reference
for each- This point is called the origin.
We select a unit of distance for x and a unit of. distance for y
which may or may not be the same, according to the problem
under discussion. As before, we represent positive numbers on
the X axis to the right, and negative numbers to the left. Positive
values of y are represented above the X axis, and negative values
below it. The arrowhead on the
axes indicates the positive direc-
tion. Any pair of values of x
and y, written (x, y), may now
be represented by a point on
the plane which is x units from
the Y axis and y units from the
.Y axis. Thus if a; = 0, y = 0,
written (0, 0), the point repre-
sented is the origin. The point
(3, 0), i.e. a; = 3, y = 0, is found
by going three units of x to the
right, i.e. in the positive direction of x and no units up. The point
(4, 3), i.e. a; = 4, y = 3, is found by going four units of x to the
right and three units of y up. The point (— 3, 1) is found by
^11
-3, 4)
r*.
(.0,1)
U,3.
?.0I
(4,
■2)
GRAPHICAL REPRESENTATION 89
going three units of x to the left and one unit of y up. The point
(— 3, — 4) is found by going three units of x to the left and four
units of y down. In fact, if we let both x and y take on every
possible pair of real values, we have a point of our plane corre-
sponding to each pair of values of (x, y). Conversely, to every
point of the plane correspond a pair of values of (x, y). These
values are called the coordinates of the point. The value of a:,
i.e. the distance of the point from the Y axis, is called its
abscissa; the value of y, i.e. the distance of the point from the
X axis, is called its ordinate. If the point (x, y) is conceived as a
moving point, and if no restriction is placed .
on the value of the coordinates so that they
, 1 .-.,.._ 2nd Quadrant
take on every possible pair of real values, (-,+)
iBt Quadrant
AlJi Quadrant
every point in the plane is reached by the ^
moving point (x, y). srd Quadrant
The X and Y axes divide the plane into C-'-)
four parts called quadrants, which are num-
bered as in the figure. The proper signs of the coordinates of
points in each of the quadrants are also indicated.
EXERCISES
The following exercises should be carefully worked on plotting paper,
which can be bought ruled for the purpose.
1. Plot the points (2, 3), (0, 4), (- 4, 0), (- 9, - 2), (2, - 4).
2. Plot with the aid of compasses the points (l, VS), (V3, — V2),
(2 + V3, 2-V3),(-V2, -V2).
3. Plot the square three of whose vertices are at (— 1, — 1), (— 1,+ 1),
( + 1, — 1). "What are the coordinates of the fourth vertex ?
4. Plot the triangle whose vertices are (2, 1), (—6, — 2), (—4, 4).
5. Plot the two equilateral triangles two of whose vertices are (6, 1),
(— 6, 1). Pind coordinates of the remaining vertices.
6. If the values of the coordinates {x, y) of a moving point are restricted
so that both are positive and not equal to zero, where is the point still free
to move ?
7. If the coordinates (x, y) of a moving point are restricted so that con-
tinually y = 0, where is the point still free to move ?
90
QUADRATICS AND BEYOND
8. What is the abscissa of any point on the T axis ?
9. The coordinates of a variable point are restricted so that its ordinate
is always 2. Where may the point move 1
10. If both ordinate and abscissa of a point vanish, can the point move ?
Where vyill it be ?
11. Plot the quadrilateral whose vertices are (0, 0), (— 6, — 3), (5, — 5),
(—1, — 8). What kind of a quadrilateral is it?
12. The coordinates of three vertices of a parallelogram are (— 1, — 1),
(6, 2), (— 1, — 6). Find the coordinates of the fourth vertex.
13. The coordinates of two adjacent vertices of a square are ( — 1, — 2) and
(1, —2). Find the coordinates of the remaining vertices (two solutions). Plot
the figures.
14. The coordinates of two adjacent vertices of a rectangle are {— 1, — 2),
(1, — 2). What restriction is imposed on the coordinates of remaining
vertices ?
15. The coordinates of the extremities of the bases of an isosceles triangle
are (1, 6), (1, — 2). Where may the vertex lie? What restriction is imposed
on the coordinates (x, y) of the vertex ?
101. The graph of an equation. The equation x = 2 y is satis-
fied by numberless pairs of Yalues (x, y); for example, (2, 1),
(0, 0), (1, ^), (— 2, — 1) all satisfy the equation. There are, how-
ever, numberless pairs of values which do not satisfy the equation ;
for example, (1, 2), (2, -1), (-1, 1),
(0, — 1). The pairs of values which
satisfy the eqtiation may be taken
as the coordinates of points in a
plane. The totality of such points
would thus in a sense represent the
equation, for it would serve to dis-
tinguish the points whose coordi-
nates do satisfy the equation from
those whose coordinates do not.
After finding a few pairs of values which satisfy the above equar
tion we note that any point whose abscissa is twice the ordinate,
i.e. for which x = 2y, is a point whose coordinates satisfy the
equation. Any such point lies on the straight line through the
origin anji the point (2, 1). We can then say that those points
GRAPHICAL REPEESENTATION 91
and only those which are on the straight line represented in the
figure have coordinates which satisfy the equation. This line is
the graphical representation or graph of the equation.
The equation of a line or a curve is satisfied hy the co'drdinates
of every point on that line or curve.
Any point whose coordinates satisfy an equation is on the
graph of the equation.
102. Restriction to coordinates. In § 100 it was seen that a
moving point whose coordinates were unrestricted took on every
position in the plane. We now see that when the coordinates of
a point are restricted so as to satisfy a certain equation (as a; = 2 j/),
the motion of the point is no longer free, but restrained to move
along a certain path. Thus, for instance, the equation x = 4 means
that the path of the moving point is so restricted that its abscissa
is always 4. Its ordinate is still unrestricted and may have any
value. This shows that the plot of k = 4 is a straight line four
units to the right of the Y axis and parallel to it, for the abscissa
of every point on that line is 4, and every point whose abscissa is
4 lies on that line.
EXERCISES
Determine on what line the moving point is restricted to move by the
following equations. Draw the graph.
1. a; = 6. 2. E = 0. 3 . y = f .
i. x = y. 5. y = 2. 6. Sx = y.
7. 2x = y. S. y = 0. 9. a; = -f.
10. x = 3y. 11.3y=-x. U.x + y = 0.
13. 6a; = 11. 14. ^ = -3. 15.2x = -3y.
16. x = -2y. 17. a; = -1. 18.2x-6y = 0.
103. Plotting equations. Plotting an equation consists in find-
ing the line or curve the coordinates of whose points satisfy the
equation. Thus the process of § 101 was nothing else than plot-
ting the equation x = 2y. This may be done in some cases by
observing what restriction the equation imposes on the coordinates
of the moving point ; but more often we are obliged to form a
92
QUADRATICS AND BEYOND
table of Tarious solutions of tlie equation, and to form a curve
by joining the points corresponding to these solutions. This
gives us merely an approximate figure of the exact graph -which
becomes more accurate as we find the coordinates of points closer
to each other on the line or curve.
Rule. When y is alone on oTie side of the equation, set x equal
to convenient integers and compute the corresponding valttes of y.
Arrange the results in tabular form. Take correspondirig
values of x and y as coordinates and plot the various points.
Join adjacent points, m,aking the entire plot a smooth curve.
When X is alone on one side of the equation integral values of y may he assumed
and the corresponding values of x computed.
Care should be taken to join the points in the proper order so that the resulting
curve pictures the variation of y when x increases continuously through the values
assumed for it. By adjacent points we mean points corresponding to adjacent
values of x.
Any scale of units along the X and Y axes that is convenient may be adopted.
The scales should be so chosen that the portion of the curve that shows considerable
curvature may be displayed in its relation to the axes and the origin.
When there is any question regarding the position of the curve between two
integral values of x, an intermediate fractional value of x may be substituted, the
corresponding value of y found, and thus an additional point obtained to fix the
position of the curve in the vicinity in question.
EXERCISES
Plot:
1. x2-4x + 3 = y.
Solution : In this equation If we set a; = 0,
1, 2, 3, etc., we get 3, 0, — 1, 0, etc., as cor-
responding values of y. Thus the points
(0, 3), (1, 0), (2, - 1), (3, 0), etc., are on
the curve. These- points are joined in order
by a smooth curve.
y
X
y
3
-1
8
-2
15
-1
3
8
16
GRAPHICAL REPRESENTATION
93
2. y = x^-1x + \.
4. y = a;2_3a; + 2.
6. y = x'-2a;+ 1.
8. 2/ = 2i2-6a; + 7.
10. y = 23?-&x-S.
3. y = x2 + 1.
5. y = x^ — ix.
7. 2/ = x2 + 6x + 5.
9. 2/ = 2a;2_3a,4.4.
11. y = a;2-12a; + ll.
104. Plotting equations after solution. When neither x nor y
is ah-eady alone on one side of the equation, the equation should
be solved for y (or x) and the rule of the previous section applied.
It should be noted that when a root is extracted two values of
y may correspond to a single value of x.
EXERCISES
3j/2 = 9-2a;2,
2^ = ±V3-|a;2.
Assuming the various integral values for x, we obtain the following
Plot:
1. 2x2 + 3y2 = 9.
Solution
tabic
and plot:
X
y
lO
y
±V3 = ±1.7
-1
±Vl = ±1.5
1
±VI = ±l-5
-2
±VI = ± -57
S
±Vi=± -57
-3
imaginary
+ 3
imagmaiy
In this example, when x is greater than 3 or less than ~ S, y is imaginary.
Thus none of the curves is found outside a strip x = ± 3.
To find exactly where the curve crosses the X axis, the equation may be
solved for x and the value of x corresponding to j^ = found. Thus
x = ± Vl-|2/^-
Tiy = 0,x = ± Vi = ^■^- '^^^ point is included in the plot
2. xy = i.
Solation :
y = :
94
QUADRATICS AND BEYOND
Form table for integral values of x.
12
Since this table does not give us any
idea of the curves between + 1 and — 1,
we supplement the table by assuming
fractional values for x.
V
X
y
4
-1
-4
2
-2
-2
1
-3
-^
1
-4
-1
1
-6
-i
\
-8
-\
*
-12
-i
3. a;2 = 2/8.
5. xy =— 1.
7. a;2 + 3^2 = 16.
9. a;2 + 2/^ = 25.
11. 2xy + 3x = 2.
13. a;2/ + 2/2 = 10.
Hint, a; = ^■
4. a;2/ = 16.
6. aij/ = a; + 1.
8. k2-j/2 = 9.
10. a;2 + a; = 12 2/.
12. x2 + 93,2 = 36.
14. x — y + 2xy = 0.
Hint. a; = — -^-— •
1 + 22/
12
a;
y
-f
-6
-*
-8
-i
-12
15. 6x^+2x+3y^=0. 16. a;2 + 2x + 1 =y2 - 3y.
105. Graph of the linear equation. The intimate relation
between the simplest equations and the simplest curves is given
in the following theorems.
Theorem I. The graph of the equation y = ax is the straight
line through the origin and the point (1, a), where a is any real
number.
The proof falls into two parts.
First. Any point on the line through the origin and the. point
(1, a) has coordinates that satisfy the equation. Let P (Figure 1)
with coordinates (a;', y') he on the line OA. By similar triangles
1 x''
or
ax'.
GKAPHICAL REPHESENTATION
95
Thus tlie coordinates of any point on the line satisfy the equation.
''(.^•y)
Second. Any point whose coordinates satisfy the equation lies
on the line.
Let the coordinates (x', y') of the point P (Mgure 2) satisfy the
equation. Then -we have
y' = ax',
— = a.
Let the ordinate y' cut the line at B. Then by the first part
of the proof £(. _ ^,^
BC
X
a=—,'= — T' or y' = BC.
x' x'
Thus
Hence P lies on the line.
Theorem IL The graph of any linear equation in two vari-
ables is a straight line.
The general linear equation
Ax+By+C =
may be written in the form
y = ax + b,
(1)
(2)
where a =-4 and 6 =- -I, provided -B ^ (§ 7). If B = 0,
is jd
the equation Ax + C = may be put in the form
C
x=--,
QUADRATICS AND BEYOND
provided A ¥^ 0. This is evidently the equation of a straight
line parallel to the Y axis (§ 102). If 5 = and A = 0,yfe have
no term left involving the variable. Thus the only case for
which the theorem demands proof is
when B =^ 0, and the equation may be
reduced to form (2). By Theorem I
we know that the graph of y = aa; is
a straight line. If, then, we add to
every ordinate y of the line y = ax the
constant b, the locus of the extremi-
ties of the lengthened ordinates will
lie in a straight line, as one can easily
prove by Geometry. But any point (x, y) on the upper line is
such that its ordinate y is equal to the ordinate of the lower
line, i.e. ax, and in ad-dition the constant 6 ; that is, y.= ax + b.
Also, since the upper line is the loous of the extremities of the
lengthened ordinates, every point whose coordinates satisfy the
equation y= ax + b is on this upper line. Thus the equation (1)
has a straight line as its graph.
CoKOLLAEY. Two lines whose equations are in the form
y=:ax + b, (3)
y = ax + V (4)
are parallel..
For the value of the ordinates of (3) corresponding to a given
abscissa, say x-^, is obtained from the ordinate of (4) corresponding
to the same abscissa by adding the constant b — b'. Thus each
point on (3) is always found b — b' units above (below i£ b — b'
is negative) a point of (4). Thus the Hnes are parallel.
106. Method of plotting a line from its equation. Since the
equation y = ax + b is satisfied by the values (0, b), the graph cuts
the Y axis b units above (below if b is negative) the origin. Since
it is satisfied by the values (1, a + b), the graph passes through
the point reached by going one unit of x to the right of (0, b) and
a units up (down if a is negative). These two points determine
the line. We may then plot a linear equation by the following
GRAPHICAL REPRESENTATION
97
Etjle. Reduce the given equation to the form
y = ax + h.
Flat the point (0, V) as one of the two points that determine
the line.
From this point go one unit of x to the right and a units
of y up (dovm if a is negative) to find a second point that lies
on the line.
Draw the line through these two points.
EXERCISES
Plot :
1. 6a; + 22/ -5 = 0.
Write in the form
2/ = era + 6,
and we tiave
y = -3a; +
a =-3,
6 = 1.
Plot the point (0, J).
From this point go one unit of x to the right and three
units of y down to find the second point, which helps
determine the line.
2. 6x~Zy+n = (i.
y = 2is + -y.-
3. a; — ^ = 0.
5. X + y = i.
7. 2x~y = i.
9. ?,x-y = 0.
11. x-?,y = l6.
13. x = %{2-y).
15. x-y-\=0.
17. a; + y + 1 = 0.
19. 12a;-3y=l.
4. x — y — 5 = 0.
6. 2a! = 6(1 -J/).
8. 12a; + 102/ = 5.
10. lhx-Wy = i.
12. 2x + 2/ + 3 = 0.
14. 2a; -63/ -1 = 0.
16. 2x-22/-6 = 0.
18. 3a; -62/ -4 = 0.
20. 7 a; -82/ -9 = 0.
107. Solution of linear equations, and the intersection of their
graphs. The process of solving a pair of independent linear
equations consists in finding a pair of numbers (x, y) which
satisfy them both. Though each equation alone is satisfied by
countless pairs of values (x, y), we have seen that there is only one
pair that satisfy both equations. Since a pair of values which
98 QUADRATICS AND BEYOND
satisfy an equation are the coordinates of a point on its graph, it
appears that the pair of values that satisfy simultaneously two
equations are the coordinates of the point common to the graphs
of the two equations, that is, the coordinates of the points " of
intersection of the two lines.
EXERCISES
Find the solutions of the following equations algebraically. Verify the
results hy plotting and noting the coor-
dinates of the point of intersection.
3a; -42/ + 16 = 0,
Solution :
3x-iy + W =
Sit- y - 7 =
1.
Sy
-23 =
y = -¥-■
Substituting in (2),
3a; = 7 + -¥- = */-•
X = -V.
To plot (1) and (2) we get the equa-
tions in the form y = CKC -|- ft and apply
the rule. Thus
y = \x + i.
y = Sx-1.
2 2x + 3y = (
■ 7a; + 2/ = 2.
5.
3x-2y = l,
3a; + 2y = 5.
_ 2x + y = S,
■ 8x-7j/ = 1.
a; + 2/ =-4,
4a; -32/ = 5.
14.
17.
x + y = 6,
4a; -22/ = 28.
a; - 2/ = - 4,
2s + 62/ = 16.
g X + 2/ = 5,
■ 3x + 2/ = l.
g 3a; -7 2/ = 9,
■ x + 2y = S.
g 2x-52/ = 0,
X — 2^ = 3.
12.
X - 2/ = 7,
2x-82/ = 3.
jg x-2/ = 2,
ix — by =
18.
3x-|-2s^ = 9,
8x-2/ = 2.
4 x-32/ = -7,
4x + 2/ = ll.
„ x + 2/ = -7,
■ 2x-32/ = 6.
10. =" + 22/=-
2x-2/ = 0.
10.
13 ^-y = l.
2x-42/ = -16.
16.
6x — 5^ = 5,
2x + 32/ = -20.
20,
2x-|-62/ =
3x + 2^ = 2
GRAPHICAL KEPRESEISTTATION 99
108. Graphs of dependent equations. We have defined (§ 57) dependent
equations as those that are reducible to the same form on multiplying or
dividing by a constant Thus two dependent equations are reducible to the
same equation of the form y = ax + b. Hence dependent equations have as
their graphs the same straight line. We see now the geometrical meaning of
the statement that dependent equations have countless common solutions.
Since their graphs have not one but countless points in common, being the
same line, it is clear that the coordinates of these countless points will
satisfy both equations.
109. Incompatible equations. By our definition (§ 60) incompatible equa-
tions have no common solution. Since every pair of distinct lines have a
common point unless they are parallel, we can foresee the
Theorem. Incompatible equations have parallel lines as graphs.
Let the equations
ax + by + c = 0, (1)
a'x + b'y + c' = (2)
be incompatible. This is true (§ 60) when and only when
a6' - a'6 = 0. (3)
Let us then assume (3).
Write (1) and (2) in the form
2, = --^--. (4)
This may be done if neither b nor b' equals zero. If both b and 6' vanish,
the lines (1) and (2) are both parallel to the Y axis and hence to each other,
which was to be proved. But if only one of them vanishes, say 6 = 0, then
by (3) a = (§ 6), in which case (1) does not include either variable. Thus
we may assume that neither 6 nor b' vanishes and that (4) and (5) may be
obtained from (1) and (2).
By (3) E - ^'
b~V
Our equations (i) and (5) become
a c
a c'
y = — X — ,
" b b' ■ ' .
which represent parallel lines, by the Corollary, p. 96.
, This theorem completes the discussion of the graphical representation of
the possible classes (§ 61) of pairs of linear equations.. ..
■^^^^^^ QUADRATICS AND BEYOND
<>^'
.^^
EXERCISES
Plot and solve :
. 8x + 2y = 3, - 2x + 6y = l,
' 4:x + y = 8. ' x + Sy = 1.
„ 10x-5y = 15, . 2z-Sy = 6,
2x-y = S. ' 8x-12y = 2i.
, x-1y = l, g 12a; -62/ =18,
4a;-28w = 56. ■2a;-y=l.
7.
x-Sy = 2, ^ 2x-3 + y = 0,
6x-lSy = S6. '4x-T + 2y = 0.
110. Graph of the quadratic equation. Let
y = ax^ + bx + c, (1)
■where as usual a, b, and c represent integers and a is positive.
If we let X take on Tarious values, y will have corresponding
values and we may plot the equation as in § 103. A root of the
quadratic equation
ax^ + bx + c = (2)
is a number which substituted for x satisfies the equation, that
is, gives the value y = in (1). Thus the points on the graph
of (1) which represent the roots of the equation (2) are the
points for which y = 0, that is, where the curve crosses the
X axis. The numerical value of the roots is the measure of
the distance along the X axis from the origin to the points where
the curve cuts the axis. Since this distance is always a real
distance, only real roots are represented in this manner.
Thboeem. If the graph of (1) has no point in common with
the X axis, the equation (2) has imaginary roots, and conversely.
Every equation of form (2) has two roots either real or imagi-
nary (§ 89). If the graph of (1) has no point in common with
the X axis, there is no real value of x for which y = 0^ i.e. no
real root of (2). The roots must then be imaginary.
Conversely, if (2) has only imaginary roots, there is no real
value of X which satisfies it, i.e. which mates y = in (1).
Thus the curve has no point in common with the X axis.
GRAPHICAL REPRESENTATION 101
This suggests the following universal
Principle. JVon^intersection of graphs corresponds to imagi-
nary or infinite-valued solutions of equations.
HI. Form of the graph of a quadratic equation. Consider the
equation y^2x^+lx + 2. (1)
By substituting for x a very large positive or negative number,
say a; = ± 100, y is large positively. Thus for values .of x far to
the right or left the curve lies far above the X axis. If we
assign to ya. certain value, say 2/ = 2, we can find the correspond-
ing values of x by solving a quadratic equation. Thus in (1) let
'^^ ' 2 = 2x^-\-lx + 2,
or ■ 2x^ + 7 x = 0.
The roots are ajj = — 3J, x^ = 0.
Hence the points (— 3^, 2) and (0, 2) are on the curve (§ 101).
That is, if we go up two units on the Y axis, the curve is to be
found three and one half units to the left and also again on the
Y axis. If in (1) we let y = — 4, the corresponding values of x
are very nearly equal to each other (— IJ and — 2), which means
that the curve meets a line parallel to the X axis and four units
below it at points very near together. The question now arises.
Where is the bottom of the loop of the curve ? This lowest point
of the loop has as its value of y that number to which correspond
equal values of x. Hence we must determine for what value of
y the equation (1), that is, the equation
2x^ + lx+{2-y)=0,
has equal roots. Comparing with the equation ax^ + hx -\- c =
(S 98), we have „ „ ■, ^
^^ " 2 = a, 1 = h, 2 — y = c.
Thus the condition J" — 4 ac = becomes
49 - 4 • 2 (2 - 2/) = 0,
49-16 33 , ,
102
QUADRATICS AND BEYOND
Substituting this value of y in (1), we get — J as tlie corre-
sponding value of X.
This gives a single
value of y for which,
the values of x are
equal; hence the graph
of (1) is a single fes-
toon as in the, figure.
If we take the gen-
eral equation
ax^ + bx + c = y,
we find precisely similarly that the hottom of the loop is at a
point whose ordinate is
b^ — Aae A
y = -
y
X
2
-4
-H
— .3+ or — 3.2+
or - 3^
- 1| or - 2
-1*
4a
4a
Thus we see again that if the discriminant is negative the graph
is entirely above the X axis and both roots are imaginary (§§ 98,
110), since the ordinate of the lowest point of the loop is positive.
If the discriminant is positive, the graph cuts the X axis and both
roots are real.
The results of this section enable us to determine a value of
y from the coefficients which determine the lowest point of the
loop of the curve precisely, and hence to show beyond question
from the graph whether the equation has real or imaginary roots.
EXERCISES
Plot the following equations and determine by measurement the roots in
case they are real. Find in each case the lowest point on the loop.
1. x^ + x + l = y.
4. x^ + 1x + G = y.
7. x^-6x + l = y.
10. x^ + 2x-l-y.
13. 2x^-x-S = y.
2. a;2-4a; + 7 = j^.
5. a;2_6a; + 9 = y.^
8. x^^6x + 5 = y.
11. x^ — 4:x + 4i = y.
14. 3x^ + Sx+5 = y
16. What is the- characteristic feature of the plot of an equation whose,
roots are equal ?
3. a2 _ 6 a; ^ xo = y.
6. 3 a;2 - 7 X - 6 = y.
9. 2x^-9x + 7 = y.
12. Sx^-ix-3 = y.
15. 4ii;2+12a;+9 = y.
GRAPHICAL REPRESENTATION .103
112. The special quadratic ax^ + ftas = O. When in the quad-
ratio equation
ax^ -{-Ix^ = 0, (1)
c = 0, we can always factor the equation into
aa? + bx =: X (ax + &) = 0,
or a; ( a; + - I = 0.
\ «/
Thus the roots are aii = 0, Kj =
a
Conversely, if a; = is a root, then (§ 95) a; — 0, or x, is a factor
and the equation can haye no constant term.
This affords the
Theorem. A quadratic equation has a root equal to zero
when and only when the constant term vanishes.
We show in a similar manner that both roots of the equation
(1) are zero when and only when J = c = 0.
EXERCISES
1. Prove the theorem just given by considering the expressions for the
roots in terms of the coefficients (§ 89).
2. For wliat real values of k do the following equations have one root
equal to zero ?
(a) a;2 + 6a; - ft + 1 = 0. (b) 2a;2 - 3a; + 4" - 1 = 0.
(o) x2 + 6a; + k^ +.1 = 0. {&) 2x^ - ix + k^ -Zk = Q.
(e) ix'^ + 2kx-2k'^-ik-2=Q. (f) Qx'^ - ix + %k^ + k + 1 = {).
3. What is the characteristic feature of the plot of an equation which has
one root equal to zero ?
4. For what real value of J; will both roots of the following equations vanish ?
(a) - + 3a; - 1 = 0. (b) a;2 + (i^ + 3) a; + i; - 3 = 0.
(o) a;2 + (i24.1)a; + 1 = 0. (d) a;2+ (S;-3)a; + 2i;2 - 54 - 3 = 0.
(e) a;2-|-(i:-l-l)a; + ft2-l = 0, (f) (k-S)x'^+{k'^-9)x-^k'^-ik-\-Z=0.
104 QUADRATICS AND BEYOND
113. The special quadratic ax'^ + c = O. This equation may
be written in the form as' H — = and factored * immediately into
a
(.+^£)(_^.„,
which shows that the roots are equal numerically hut have oppo>
site signs. The roots are
'^^=aR' '^^^-aR-
Since in the equation ax'' -{- e = i/ the Tariahle x occurs only in
the term a;", we get the same value of y for positive and negative
values of x. Hence the loop which forms the graph of the equar
tion is symmetrical with respect to the Y axis.
114. Degeneration of the quadratic equation. The equation
ax'' + bx + c =
has the roots aji ■■
_ j + Vs^
-4:ao
2a
-h-^sfW-
- 4: ac
'^^~ 2a
We wish to find the effect on the roots Xi and x^ when a
becomes very small. If we let a approach 0, then Xi approaches
an expression of the form —> which must always be avoided.
Eationalize the numerators and we get
4 ac 2 c
2a(-S-V62_4ac) - 5 _ Vj" - 4 ac
4ac 2 c
2a{-b + \'b''-4:ac) -b+y/b^-iac
As a approaches 0, evidently 6" — 4 ac approaches P, Xi ap-
proaches — 7> and cca, since its denominator becomes very small,
iai
Imaginary (§ 152),
GRAPHICAL REPEESENTATION
105
increases without limit, that is, approaches infinity. Thus the
quadratic equation approaches a linear equation when a approaches
0, and one of its roots disappears since it has increased in value
beyond any finite limit. The loop-shaped graph of the quadratic
equation must then approach a straight line as a limit when a
approaches 0. This is made clear from the following figure, where
a has the successive values 1, \, ^, ^, 0.
In the figure the curves represent the following equations :
x'-
-1-2=..
(I)
x^
5
-i-2-»-
TO
x^
10"
i-.-.
(III)
x^
50"
-1-^-,.
(IV)
-i-.=.
(V)
^
\
1
K
^
—
j_
—
—
—
1^
-
\
\
\
1
/
7
\
\
1
>^
s.
h
1
/
I
T,
"
X
^
\
^
i
/
/
f
"^
^
^
/
/'
/
/
X
N*
V
A
'
y
/
">
^
—
''
-*!
^
■ — ,
■^
^
-^
IV
-
—
—
—
—
—
^
^
^^
■Y
In a similar manner we can show that when in the equation
bx + c = (i,b approaches as a limit, the root of the linear equa-
tion becomes infinite.
EXERCISES
1. What real values must k approach as a limit in order that cue root of
each of the following equations may become infinite ?
(a) fcr2-|-6a;-|-l = 0. (b) (J;2 + 1) x^ -|- a; -|- 1 = 0.
(o) (kx-\Y-{x + 2y = (k + xY. (d) 42-)- 4 42x2 -(a -1)2 -I- 2 = 0.
(e) V2fa;-l-t-V6J:-l = Vfci;-l-l. (f) Va - A; -f VxTk = Vfca; -f- 1.
X ■
(g)^^ + ^±i + £ = o.
^^' 4 - 1 X + 1 fc2
«>ifi-xi5=Nf
'44-1
fc2
^-1)2 (4-^_l^
^ ' (X -I- 1)2 ^ (ft -I- 1) k
(j) (ft2 _ i)x2 -(- (fc - l)x -f *;2 -H 4 fc - 6 = 0.
106 QUADRATICS AND BEYOND
2. What real values must k and m, approach as a limit in order that both
roots of the following may become infinite ?
(a) &c2 + ma; + 1 = 0.
(b) (2fc-m)x2 + fcc-2 = 0.
(c) 21a? + (3m - 1 + k)x = Sx^ - 1.
(d) {k - l)a;2 + (fe + m + l)x + 3 = 0.
(e) s2 _ a; - 2(fc + m)x = (k + m)(7?- 1).
(f) (fc + m)^ + 2(fc + m) + 1 = a;2 - 2s.
(g) (k + m + l)a;2 + (2 ft - m - l)a; + 1 = 0.
(h) (2i; + m + 2)x2 + (4fc + 2m + 8)a; + 3 = 0.
115. Sum and difference of roots. Let x^ and xj be the roots of
x'' + bx + c = <d. (1)
Then (§ 95) a; — ajj and x — x,, are factors, and their product
x^ — (xi + ccj) a; + x^x^ is exactly the left-hand member of (1).
ConsecLuently the equation
x^ + bx + c = x^ —(xi + Xi) X + X1X2
is true for all values of x. Hence by § 96
- {X, + X^)=:,b, (2)
X1X2 = 0. (3)
We may state these facts in the
Theorem. The coefficient of x in the equation x^ + hx+c—O*
is equal to the sum of its roots with their signs changed.
The constant term is equal to the product of the roots.
EXERCISES
1. Prove the statement just made from the expression for the roots in
terms of the coefficients (§ 89).
2. Form the equations whose roots are the following :
(a) 6,1. (b)hh' (c)f,3. (d)-^, -6.
(e) i, i- (i) -h+h (g) 2 + V3,-2 - V3. (h) - V3, V3.
* "We should for the present exclude the case where 6* — 4 c < 0, since the roots x^ and
x^ are then imaginary and we have not as yet defined what we mean by the sum or
the product of imaginary numbers. We shall see later that the theorem is also true in
this case.
GRAPHICAL REPRESENTATION 107
3. If 4 is one root of a;^ _ 3 ^ + c = 0, what value must c have ?
Solution : Let Xi be the remaining root.
Then by (1) ~(xi + i) = - 3,
or xi= — 1.
By (2) c = xi-i = (-l)i = -i.
4. I'ind the value of the literal coefficients in the following equations.
(a) !c3 + 61 — 9 = 0. One root is 3.
(b) x^ + lx + c = 0. One root is 2.
(c) 0x2 + 3a; - 4 = 0. One root is 2.
(d) 0x2 + 3x + 4 = 0. One root is 7.
(e) aa;2 + 2 a; + 6 = 0. One root is 6.
(f ) a;2 + 6a; + 4 = 0. One root is — 1.
(g) a;2 — 6a; - 6 = 0. One root is — 3.
(h) a;" + 5a; + 6 = 0. One root is — 6.
(i) 2 12 — 6 a; — c = 0. One root is — 4.
(i) x^ — 6 X + c = 0. One root is double the other,
(k) x" + c = 0. The difference between the roots is 8.
(1) a;2 — 5 a; + c = 0. One root exceeds the other by 3.
(m) x^ — 7x + c = 0. The difference between the roots is 6.
(n) x^ — 6x + c = b. The difference between the roots is 4.
(0) a;2 — 3 a; + c = 0. The difference between the roots is 2.
(p) x" — 2x + c = 0. The difierence between the roots is 8.
116. Variation in sign of a quadratic. It is often necessary to know the
sign of the expression
^ aa;2 + 6a; + c
for certain real values of x, and to determine the limits between which x may
vary while the expression preserves the same sign. We assume as usual that
a is positive.
Theorem I.* If the discriminant of ax^ + 6a; + c is positive, the quadratic
is negative for all values of x between the values of the roots of the equation. For
other values of x (eccepting the roots) the quadratic is positive.
* If a "were negative, Theorem I would read as follows : 1/ the discriminant is posi-
tive, the quadratic is positive for all values of x between the values of the roots of the
equation. For other values of x (excepti/ng the roots) the quadratic is negative.
Wlien a is negative Theorems IX and 111 may lae modified in an analogous manner.
108
QUADRATICS AND BEYOND
In § 98 we found that when the discriminant of a quadratic equation is
positive the equation has two real roots. If two roots are real, the loop of
the graph of the equation (ufi + 6x + c = y cuts the X axis in two points
(§ 110) as in the figure. The roots are
represented by A and B, and any real
value of X between the roots is repre-
sented by a point P in the line AB.
Since the curve is below the X axis at
any such point, the value of y, i.e. of
the expression ax^ + 6x + c for values
of X between the roots, is negative.
The value of the expression for any
value of X greater or less than both
roots is seen to be positive, since for
such points, for example Q and JB, the graph is above the X axis.
Theorem II. If the discHminant of ax^ -\-bx + c is negative, the expression
is positive for all real values of x.
When the discriminant is negative the entire graph of ax^ + bx +'c = y is
above the X axis {§ 111), and consequently for any real value of x the corre-
sponding value of y, i.e. the value of ax^ -|- 6a; -|- c, is positive.
Theorem III. If the discriminant of ax^ + bx + cis zero, the value of the
expression is positive for all values of x except the roots of the equation
ax^+bx + c = 0.
Hint. See example 16, p. 102.
We may restate these three theorems and prove them algebraically as
follows :
Theorem IV. If the discriminant of the quadratic ax^ + bx + c is positive,
the values of the quadratic and a differ in sign for all values ofx lying between
the roots, and agree for other values.
If the discriminant is zero or negative, the value of the quadratic always
agrees with a in sign.
Case I. Since the discriminant is positive, the equation ax^+bx+c=0
has two unequal real roots, as xi and x^, of which we wiU assume xi is the
greater, and we may write the quadratic in the form
ox^ + 6x + c = a (x — Xi) (x — X2).
Now for any value of x between Xi and X2 the factor x — xi is negative,
while X — X2 is positive, which shows that the quadratic is opposite in sign
to a for such values of x. For other values of x both these factors are either
positive or negative, and for such values the quadratic is of the same sign
as a.
GKAPIilCAL KEPRESENTATION 109
Case II. Since the discriminant V' — iac is negative and tlie roots are
**i, * 6 ± V62 - 4 ac .^ ^, , ^.
of tlie form , we may write the quadratic
2a
\ 2a 2a /\ 2a 2a /
r/ 6 \2 62_4ac"l*
= "L(" + 2-^)-^^J
L\ 2o/ 4a2 J
Now for any value of x the expression (a; H 1 is positive, and since
6^ — 4 ac is negative, 4 ac — 6'' is positive ; and we observe that the last
member of the equation has tlie same sign as a.
Case III. Since the discriminant is zero, the roots are equal and the
expression has the form
OJ? + 10, + c = a{x — Xif,
which has evidently the same sign as a, for any value of x.
EXERCISES
1. Between what values of x is the expression Va;^ — 6 1 + 4 imaginary ?
Solution : The roots ofa;^ — 5a; + 4 = are 4 and 1.
The discriminant A = 6'' — 4 ac = 25 — 16 = 9is positive.
Thus by Theorem I or IV, if 1 < a; < 4 f the expression under the radical
sign is negative and the whole expression is imaginary.
2. For what values of k are the roots of
(4 + 3)a;2 + fcc + l=0 (1)
(a) real and unequal ? (b) imaginary ?
Solution : a = k + S, b = k, c = 1.
A = 62 - 4oc = 42 - 4(ft + 3) = 42 - 4fc - 12.
(a) If A > 0, the roots of (1) are real and unequal.
The roots of A;^ - 44 - 12 are * = - 2 and 6
Then, by Theorem II, if
fc< -2or A:>6, A>0.
(b) By Theorem I, if - 2 < ft < 6, A < 0,
and the roots of (1) are imaginaiy.
* Bee § 1S2. t Kead " 1 is less than x which is less than 4 " or " a is between 1 and 4."
110 QUADRATICS AND BEYOND
3. Determine for what values of x the following expressions are negative,
(a) x^ + 2x- 1. (b) s" - 6a; + 4.
(c) z2 - 11a; + 10. (d) z^ - 16a; + 60.
(e) - !» - 2z + 1. (f) - z2 + 7z + 30.
4. Determine for what values of k the roots of the following equations
are (a) real and unequal, (b) imaginary.
(a) 3fcc2-4z-2 = 0. [b) 3^ + ikx + k^ + 1 = 0.
(c) x2 + 9kx + 6fe + J = 0. (d) z2 + (3 fc + i)x + 1 = 0.
(e) (42 + 3)a;2 + fci - 4 = 0. (f) 2x2 - 4z - 2i: + 3 = 0.
(g) kx^ + (ik+l)x + ik-S = 0. (h) (A; - l)z2 + 5fca; + 6i + 4 = 0.
(i) (fe - l)a;2 + (2fc +l)x + fc + 3 = 0.
CHAPTEE X
SIMULTANEOUS QUADRATIC EQUATIONS IN TWO
VARIABLES
117. Solution of simultaneous quadratics. A single equation
in two variables, as a;^ + ?/^ = 6, is satisfied by many pairs of
values, as (1, 2), (Vf; Vi)j (2, 1), and so on, though there are
at the same time numberless pairs of values that do not satisfy
it, as (0, 1), (1, 1), (2, 3). Thus the condition that (x, y) satisfy a
single quadratic equation imposes a considerable restriction on
the values that x and y may assume. If -we further restrict the
value of the pair of numbers (x, y) so that they also satisfy a
second equation, the number of solutions is still further limited.
The problem of solving two simultaneous equations consists in
finding the pairs of numbers that satisfy them both.
118. Solution by substitution. In this method of solution the
restriction imposed on (x, y) by one equation is imposed on the
variables in the other equation by substitution.
Example. Solve 2x^ + y^ = l, (1)
x-y = l. (2)
Solution : Equation (2) states that x = 1 + y. Thus our desired solution is
such a pair of numbers that (1) is satisfied and at the same time x is equal
toy + 1.
If we substitute in (1) 1 + y for x, we are imposing on its solution the
restriction implied by (2).
Thus 2{l + yf + y^ = l,
or 3 3/2 + 4 3/ + 1 = 0.
The roots are y = — 1, y =— ^.
Corresponding to j/ = — 1 we get from (2) a; = 0.
Corresponding to 2/ = — -^ we get from (2) a; = |.
Thus the solutions are (0, — 1) and (|, — \).
Ill
112 QUADRATICS ANB BEYOND
EXERCISES
Solve the following :
. x + y = 5, n x-y = 5, „ x + y = a,
xy = 4. xy = 36. ' x^ + i/^ = bxy.
. 3a-j/ = 5, . x + y = a, g s^ + ys = 50, ,
xy - X = 0. ■ x2 4- 2,2 = 6. ' 9x + 7 y = 70.
x2 + 2/2 = 40, 2x-3y = 4, g xy = 12,
x-32/ = 0. x2-y2 = 0. ■2x + 32/ = 18.
x:y = 9:4, jj x2:yi' = a2:62, ^^ 5x^ + y = 3xy,
s:12 = 12:2/. ' o - X = 6 - y. '2x — 2/ = 0.
13 x^-xy + y^ = 1, j^ (a; + y)(» -2y) = 7,
2x-3y = 0. ■ x-y = 3.
„ 3x2-4y = 5x-2y2^ ,g x^ + y = y'' + x-18,
3x+42/ = 10. ■ x:v = 2:3.
J- ax-by = cy, ^g x^ + 2xy + yi' = 7(x - y),
• a2x2 _ 62y2 = acxy + mi'. ' 2x — y = 5.
0x2 + (a - 6)X2/ - 6y2 = c2, 2x2-5xy+y2+10x+12y = 100,
(x + y):(x-y) = a;6. " 2x-3y = l.
0*2 J_ 0/2 ^ f]2
^^_7(x + 5)2-9(y + 4)2 = 118, 22_ ^ +^ '
X — y = 1. - = —
y n
x" + y + 1 ^ 3
24. j,2 + a; + X 2'
X — y = 1.
1 + X + X^!
x2 + y2 = 130,
23.
x + y_g_
x-y
2x-y + l 8
25.
x-2y + l 3'
x2-3xy + y2 = 5.
27.
xy + 72 = 6(2x + y),
X 2
y 3
4x+y-l 4x+y-12
29.
2x+y-l 2x+y-12
3x + y = 13.
31.
^' + ' -5,
x+2 y-l
2 4
x-1 y
26. l + y + y2
X + y = 6.
= 3,
3x-2 ^ y_g
28. y + 6 x
X — y = 4.
x(x — y) — 6y = 6
3«-^ = 3J.
x — y
^ + ? = 3
32. X y '
l) = 2(x + l).
SIMULTANEOUS QUADRATIC EQUATIONS 113
119. Number of solutions. We have proved (p. 42) that two
linear equations have in general one and only one solution.
Theorem. A quadratic equation and a linear equation have
in cfeneral two and only two solutions.
If the linear equation is solved for one variable, say x, and this
is substituted in the quadratic equation, we get a quadratic equa^
tion to determine all possible values of the other variable (i.e. y),
which must in general be two in number (§ 98). To each one of
these values of y will correspond one and only one value of x,
thus affording two solutions of the pair of equations.
EXERCISES
1. 'When may, as a special case, a quadratic and a linear equation
have only one solution?
2. When may a quadratic and a linear equation have imaginary
solutions ?
3. Find the real values of k for which the following equations have
(1) only one solution, (2) imaginary solutions.
X-^ + y^=ie, (1)
^'^' K - S- = fc. (2)
Solution : x = y -\-k.
Substitute in (1), (y + ky + y^ = 16.
or 22/2 + 2fc2^ + 42 -16 = 0.
As in § 98, a = 2 ; & = 2 & ; c = 42 - 16.
Hence A = &« - 4ac = 4*2 - 8i;2 + 128 = - 4A;2 + 128.
(1) A = when 442 = 128,
or fc = ± 4 v^. There is then only one solution.
(2) A < when 42 > 32. The solution is then imaginary.
,., X2/ = l, x + ky=.!o,
^ ' x + y = k. ^ ' a;2 + y2 = 5.
X - l>y = 1. ^a;2 — 4^ = 0.
x^-y^ = 8, a;2 + 2/2 = 26,
x + 2/ = fc. ' ix — Zy = k.
(d)^'=^2''
a=2_^. = 9,
^^' x-2y = k.
2.2 + 32/2 =
^" x-ky = l.
6,
114 QUADRATICS AND BEYOND
120. Solution when neither equation is linear. In the exam-
ples previously given one equation has been linear and the other
quadratic in one or both variables. Often when neither of the
original equations is linear a pair of equivalent (p. 41) equations
one or both of which are linear may be found. These latter equa-
tions may be solved by substitution.
EXERCISES
Solve the following equations.
When neither equation is linear, we can often obtain by addition an
equation from which by the extraction of the square root a linear equation
may be found.
1. a;2 + 2/2 = 17, (1)
xy = 4 (2)
Solution: x^ + j/^ = 17 (3)
Multiply (2) by 2, izy = 8 (4)
Add x^ + 2zy + y^ = 'it
Extract the square root, a; 4- V = ± 5- (S)
Subtract (4) from (3), x^ -ixy + y^ = Q.
Extract the square root, a; — 2/ = ± 3. (6)
Solve (5) and (6) as simultaneous equations,
a; + 2/ = ± 5,
x-y = ±Z.
x = +i, +1, -1, -4.
y = + \, +4, -4, -1.
Thus the solutions are four In number, (4, 1), (1, 4), (—1, — 4), (— 4, —1).
The following exercise affords another case where a linear equation may
be found by addition and extraction of the square root.
2. x-i + xy = 6, (1)
xy-\-y'^ = 10. (2)
Solution : Add (1) and (2),
3l^ + 2xy + y^ = 16.
Extract the square root, x + 2^ = ± 4.
Substitute In (1), x^ + x(± 4 -x) = 6,
x2±4x-x2 = 6,
x = ±f = ±lj.
Substitute in (4), y = 2^, - 2J.
Thus our solutions are (— |, — 2J), (f, + 2^).
SIMULTANEOUS QUADRATIC EQUATIONS 115
When neither of the original equations is quadratic, we can often find by
division an equivalent pair of equations one of which is linear and the other
quadratic, as in the following exercise.
3. x» + y' = 12, (1)
x + y = 2. (2)
Solution: Divide (1) by (2),
x,'i-xy + y^ ='6. (3)
Square (2), x^ + 2xy + y'=:4:
Subtract, -Sxy~2
xy^-l. (4)
Solve (4) with (2) by substitution.
When the sum of the exponents of the variables is the same in every
term, the equation is called homogeneous.
Thus, ^ + xy = 0, 2a:2y-3a!2/2-4a;3-3!/8 = 0.
When one equation is homogeneous and the other either linear or quadratic
we may solve them as follows :
4. 6a;2-7a;2/ + 2t/2 = 0, (1)
a;2 - sr = 4. (2)
Solution : Divide (1) by y^,
Let- = z,* 6«2-7z + 2 = 0.
y
Solve for z, z = J or |.
™, X 1 X 2
Thus - =- or - = -.
y 2 y Z
Solve (2) with 2x = y and 3 a; = 2 ^.
When both equations are homogeneous except for a constant term we may
solve as follows :
5. 3fi-xy + 2y^ = i, (1)
2a;2-3a;y-22/2 = 6. (2)
Solution : Eliminate the constant term by multiplying (1) by 3 and (2) by 2,
Zx^-Sxy+ 6y2 = 12, (3)
ix^-Qxy- 4y2 = 12 (4)
Subtract (3) from (4), x^-Zxy -Wifi=
* We observe that y ^0. For if y = were a value tliat satisfies equation (1), a; =
would correspond. But (0, 0) does not satisfy (2); thus y = is not a value that can occur
in the solutions of the equations.
116 QUADRATICS AND BEYOND
Divide by t/^ and let - = z, where y ^Q,
y
z2-3z-10 = 0. (5)
Factor, (z - 5) (z + 2) = 0.
The roots are - = 5, - = -2. (6)
y y
Solve (6) with (1).
When one equation is quadratic in a binomial expression we may solve as
follows :
6.
x-y-
- Va; - y = 2,
a;3 _ 2,8 = 2044.
Solution : Let
■y/x — y = z.
Then (1) becomes
z2 - z = 2.
Solving for z,
z = 2 or — 1.
Thus '
X — y =ii'
x-y=l/
3.
or
Solve (3) with (2) as in exercise
7. x^ + y^ = xy = x
+ y-
8. a;8 + 2/3 = 7 xy =
g x^ + xy^ = 2.
xy^ = 6.
x(y-l) = W,
■ y{x~V) = \%
j2 x^ + y^ = a,
xy = h.
jg x^y + xifi = a,
x^y — xy^ = 6.
,. a; + X2/ = 35,
■ y + a;2, = 32.
15 xixo + y')^!,
y(x» + y^) = l.
2x2 _ 32,2 = 6,
■ 3x2-22/2 = 19.
,- 5x2 + 2 2/2 = 22,
3x2-52/2 = 7.
(1)
(2)
(3)
j^g x2 + xy + s^2 = 2, 20 ^"'"^^ + ^ = ^'
■ x2-X2/ + 2/2 = 6. • a;2 + j;2, + j,2 = 7.
Hint. Eliminata x2 or j/2 as if k i q — o
the equations were linear equa- 21. + » + — »>
tions in x2 and 2^2. X2/ = 2x-2/ + 9.
22 x2-x2/ + 2/2 = 37, (x + 2/)(8-x) = 10,
• x2-j/2 = 40. * • (x + 2/)(5-2/) = 20.
24 (!i;2 + y2)(x + 2/)=5, 25 (x + 2/)2 = 3x2 -2,
■ a;2/{x + 2/) = - 2. " (x - 3/)2 = 8 j/2 - 11.
SIMULTANEOUS QUADRATIC EQUATIONS
117
26.
28.
30.
32.
34.
36.
38.
40.
42.
44.
« + -s/Wy = a,
V + Vxy^ = 6.
x + y = 68,
■s/x + -Vy = 10.
xi _ x^y^ + y* = VS,
x + y = 3.
4a;2-9y2 = o,
4:X^ + y^ = 8(x + y).
3xy-2{x + y) = 28,
2xy-3(x + y) = 2.
x^ + y^ + x + y = 18,
x^ — y^ + X — y = 6.
3x'-2y^ = 6(x-y),
xy = 0.
x^-xy + y'^ = 13(x-y),
xy = 12.
■Vx(l-y) +-Vy(l - X) = a,
X + y = b.
Vl^T^a Vrr^ + xy = a,
x-y = U-
E_|/_ 16
46. y x~ 15'
3a;2 + 5^/2 = 120.
xVx + yVy _ 1
48. X Vx — y -s/y 2
a;3 _ 8 = 8 - y'.
■sly — Va — a; = -^/y — x,
50. Vy — X + Va — X _ 5
~2'
52.
Va-
1 , 1_3
- + - — -1
K V 2
1+1=1
x^ y"^ 4
27.
29.
31.
33.
35.
37.
39.
41.
43.
45.
47.
49.
51.
53.
as Vx + 2/ = 3,
yy/x + y = 1.
Vx+Vy = a,
x + y = b.
Vx— Vy = 2,
(x + y)Vxy = 510.
Va;-5 + Vi7T2 = 5,
a: + 2^ = 16.
a;i/ + asr^ = a;2 + j/^,
a;!/ — a;2/-i = 2 (x^ + 3^2).
a;-2 + 22/-2 = 12,
x-2 — x-^y-^ + y-2 = 4.
x' + y^-6{x + y) = S,
x^ + y^~3(x + y) = 2S.
2x^-3xy + 6y-5z=0,
(x-2)(y-l) = 0.
V5-3cc + x2 + V6-3V + 2/2 = 6,
X + y = 3.
1 1 ^
- + - = 5,
a; y '
X — y = .3.
X y _26
y'^x~W
x^-y = 28.
f 6x jx + y _
yix + y "V 5x
xy - (x + y) = 1.
^ + ?^ = 2,
a2 62
to + a)/ _ m
bx — ay n
y X 2 '
X _^_3
2^ X 2
3V2
Hint. Let - = «, - = «.
X ' 2^
118 QUADRATICS AND BEYOND
54. ) ^l 55. 2'-! ^-1
.(l + |) = 3.
a' - 1 _ gs - 1
2/8-1 ~ 6' -l'
56. ^ + ^ 57 ^'^
y = V V 1 — x^ =
" i x-^1 " 221
i + i = ^, 1-5 = 1
„ a; 2/ 6' KQ * 2' '
1 +-l- = i. '-^-^ = 2.
a; + 2 2^ + 1 2 x + 3 y-1
60.
\ x + y/ \ x-yl '
!C2/ = 2.
PROBLEMS
1. Two numters are in the ratio 5 : 3. Tlieir product is 735. What are
the numbers ?
2. Divide the number 100 into two parts such that the sum of their
squares is 5882.
3. The sum of the squares of two numbers increased by the first is
205 ; if increased by the second the result is 200. What are the numbere ?
4. The diagonal of a rectangle is 85 feet long. If each side were longer
by 2 feet, the area would be increased by 230 square feet. Find the length
of the sides.
5. The diagonal of a rectangle is 89 feet long. If each side of the rec-
tangle were 3 feet shorter the diagonal would be 85 feet long. How long
are the sides ?
6. The sum of two numbers is 30. If one decreases the first by 3 and
the second by 2 the sum of the reciprocals of the diminished numbers is \.
What are the numbers ?
7. The sum of the squares of two numbers is 370. If the first were
increased by 1 and the second by 3, the sum of the squares would be 500.
What are the numbers ?
8. A number of persons stop at an inn, and the bill for the entire party
is |24. If there had been 3 more in the party, the bill would have been
$33. How many were in the party and how much did each pay ?
SIMULTANEOUS QUADRATIC EQUATIONS 119
9. A fruit seller gets $2 for his stock of oranges. If his stock had con-
tained 20 more and he had charged f of a cent more for each, he would have
received $3 for his stock. How many oranges had he and how much did he
get apiece for them ?
10. A man has a rectangular plot of ground whose area is 1250 square
feet. Its length is twice its breadth. He wishes to divide the plot into a
rectangular flower bed, surrounded by a path of uniform breadth, so that
the bed and the path may have equal areas. Find the width of the path.
11. In going 7500 yards one of the front wheels of a carriage makes 1000
more revolutions than one of the rear wheels. If the wheels were each a yard
greater in circumference, the front wheel would make 625 more revolutions
than the rear wheel. What is the circumference of the wheels ?
12. A man has |539 to spend for sheep. He wishes to keep 14 of the
flock that he buys, but to sell the remainder at a gain of $2 per head.
This he does and gains |28. How many sheep did he buy and at what
price each ?
13. A man buys two kinds of cloth, brown and black. The brown costs
25 cents a yard less than the black, and he gets 2 yards less of it. He
spends $28 for the black cloth and $25 for the brown. How much was each
a yard and how many yards of each did he get ?
14. A man left an estate of $54,000 to be divided among 8 persons, namely,
his sons and his nephews. His children together receive twice as much as
his nephews, and each one of his children receives $8400 more than each one
of his nephews. How many sons and how many nephews were there ?
15. A and B buy cloth. B gives $9 more for 60 yards than A does for
45 yards ; also B gets one yard more for $9 than A does. How much does
each pay?
16. A sum of money and its interest amount to $22,781 at the end of a
year. If the sum had been greater by $200 and the interest J of 1 per
cent higher, the amount at the end of the year would have been $23,045.
What was the sum of money and what was the interest ?
17. If one divides a number with two digits by the product of its digits,
the result is 3. Invert the order of the digits and the resulting number is in
the ratio 7 : 4 to the original number. What is the number ?
18. What number of two digits is 4 less than the sum of the squares of
its digits and 5 greater than twice their product ?
19. Increase the numerator of a fraction by 6 and diminish the denomi-
nator by 2, and the new fraction is twice as great as the original fraction.
Increase the numerator by 3 and decrease the denominator by the same, and
the fraction goes into its reciprocal. What is the fraction ?
120 QUADRATICS AND BEYOND
121. Equivalence of pairs of equations. In the theorems of
this section the capital letters represent polynomials in x and y,
and the small letters represent numbers not equal to zero.
Theoeem I. The 'pairs of equations
are equivalent.
If (xi, 2/1) be a pair of values that satisfy (1), then when x and
y in B^ are replaced by x-i and y^ the equation B^ = 6'' is a nimier-
ical identity. These values (x^, 3/1) must then satisfy one of the
equations fi = ± S, for if they did not, but only satisfied the equa-
tion say B = c when c =f= ±h, then the hypothesis that B^ = b^ is
satisfied by (xi, yC) would be contradicted.
Conversely, any pair of values that satisfy B = ±b evidently
satisfy B^ = ¥.
This theorem is used, for instance, in exercise 2, p. 114, and justifies the
assumption that
x^ + xy = % 1 x^ + xy = &\
a:2 + 2a;y + 2/2 = 16j *""* z + y = ±i:}
are equivalent pairs of equations.
Theorem II. The pairs of equations
,t = M(l) and t^fl(2)
AB = aly ' B = iy '
are equivalent.
\1 A = a and B = h are satisfied by a pair of numbers (x^, yi),
we multiply the identities and obtain AB = ab.
Conversely, if A = a, AB = ah are identically satisfied by a
pair (ajj, y^, since a =?!= we can divide the second identity by
the first and obtain B = b. Thus if {x^, y-^ satisfy one pair of
equations they satisfy the other pair.
This theorem is assumed in exercise 3, p. 115, to show that
«a + 3,»=12| =«^-»=2' + !'"=6\ are equivalent.
x+y=2 J x+y=2 J
SIMULTANEOUS QUADEATIC EQUATIONS 121
Theoeem III. The pairs of equations
are equivalent where a, b, c, and d are mtmbers such that
ad — hc'^ 0.
If (xi, 2/1) satisfy (1), evidently they also satisfy (2). Thus all
solutions of (1) are among those of (2).
Conversely, if (x^, y^ satisfy (2), then
a
Thus {ad-bc)B = 0.
Thus since (ad — be) 4=- 0,
£ = 0.
Similarly, .4 = 0.
This theorem has been assumed in exercises 1, 2, 3, 6, p. 114. In 1, for example,
it is necessary to show that
»=' + 2''="\(l) and «^ + 2«l/ + 2/^ = 25^(2)
ic.y = i P' a;2-2a;y + y2=9 P'
are equivalent. In this case o = c = 1, & = — d = 2. Thus ad — 6c = — 4 7^ 0.
122. Incompatible equations. When a pair of simultaneous
equations can be proven equivalent to a pair of equations which
contradict each other or are absurd, they are incompatible and
have no finite solution.
Example 1.
Subtract,
Example 2. y? ■\-y^ = 4, (1)
(2)
Multiply (1) by 4,
Subtract,
a;y =
1
x,v =
-1
=
2
a;2 + 2/2 =
4,
4a;2 + 4j,2 =
:4g.
4x2 + 42/2-
16
4a;2 + 43,2 =
49
=
33
122
QUADRATICS AND BEYOND
123. Graphical representation of simultaneous quadratic equa-
tions. Every equation that -we have considered may be rep-
resented graphically by plotting
in accordance with the method
already given (p. 93).
The solution of simultaneous
equations is represented by the
points of intersection of the cor-
responding graphs.
Thus the equations
x'^y'^^ 25,
'2,xy = 9>
have the solutions
= ±2±
' y==F2±
Vm
2 ' ^ ' -- 2
or SB = 4.9, .9, - .9, - 4.9,
y = .9, 4.9, - 4.9, - .9.
These equations have as their graph the preceding figure.
The equations
a;= + 2a;-|-4y-fl = 0,
a;-t-2y-F4 = 0,
which have the solutions
a!=± V7=±2.6,
y = -2TVi=-3.3or-.7,
have as their graph the figure shown
above.
As in the case of linear equations,
incompatible equations afford graphs
which do not intersect. Thus the graph
of the equations in example 2, p. 121, is
found to be two concentric circles, as
is shown in the adjacent figure.
■ Ill I I \y>' \ I I
_^?^^S 5
■
_ -
~- ^^^
^""^
4 A
^\ v
t
^ i^
-X ^°
^ V"
-^^^-
J T
^^
^7
SIMULTANEOUS QUADRATIC EQUATIONS 123
Simultaneous equations which have
imaginary solutions also lead to non-inter-
secting graphs (p. 101).
Thus the equations
a;» + 2/2 = 4,
have the adjacent figure as their graph.
-^n
v^
V
■ /'"'vIXI 1 1
4--^^
- —
1 — S^
^^^/i 1 1 1 r
.
EXERCISES
1. Inteipret the graphical meaning of equivalent pairs of equations.
2. Plot and solve x^ + y^ = 2,
x + y = 2.
What general statement concerning the graphical meaning of a single
solution of quadratic and linear equations does this example suggest ?
3. Plot and solve the following :
(a)
x^ + y^ = 25,
ix^ + 9y'' = Ui.
(c)
(b)
x2 + y2 = 25,
a;2 + 2/2 = 25,
5a;2-|- 2/2 = 25.
What general statement concerning the graphical interpretation of four,
three, or two real solutions of equations do these examples suggest ?
4. State the algebraical condition under which two quadratic equations
have four, three, two, or one real solutions (see p. 113).
5. Plot and solve the following :
(a)
x'' + y = 0,
a;2-33/ = 0.
ix-2y = 3,
xy — y = 0.
(fc)
(d)
32 -I- 2/2 = 9,
x^-y^ = 0.
xy = l,
a;2 + j/2 = 16.
4a;2 + 92/2 = 36,
(^^rj^'
2/ = 18.
124. Graphical meaning of homogeneous equations. Consider for example
the homogeneous equation
Sx^-10xy-Sy^ = 0. (1)
If we let z = ^, we get S-Wz-Sz'- = 0,
or • 8z2 + 10z-3 = 0,
or (4z-l)(2z + 8) =0.
124
QUADRATICS AND BEYOND
The roots are « = i and a = — f .
Thus ^ = 1 and ?^ = -?,
a; 4 X 2
iy-x = and 3x + 2y = 0.
These equations represent two straight lines through the origin which
taken together form the graph of equation (1). This
example may obviously he generalized : Any homo-
geneous equation of the form aa^ + hxy + ey^ = with
positive discriminant represents two straight lines
through the origin. Such an equation is equivalent
to two linear equations.
In an example like 5, p. 115, we obtain in place
of the given pair of equations a pair of equivalent
equations one of which is homogeneous and the other
of which is factorable. We can learn the graphical
meaning of this method of solution by studying a
particular case. Consider for example the equations :
x^ + 2xy + 1y'^ = 2i, (1)
2x''-xy-y^ = 8. (2)
By eliminating the constant terms we obtain the product of the two
equations x + y = and x — 2y = 0. Thus the problem of solving (1) and
(2) is replaced by that of solving the two following pairs of equations :
a;2 + 2xy + 7y2 = 24,
x + y = 0,
or x'' + 2xy + 'ly' = 2i,
x-2y = 0.
The graphical meaning of this .
method of solving the equations (1)
and (2) is seen in the fact that the
problem of finding the points of inter-
section of the graph of equation (1) '
with that of (2) is changed to that of
finding the intersection of the graph of (1) with a pair of straight lines.
This appears in the figure where the curves and lines are numbered as
above. The closed curve represents (1).
CHAPTEE XI
MATHEMATICAL INDUCTION
125. General statement. Many theorems are capable of direct
and simple proof in special cases, while for the general case a
direct proof is difficult and complicated.
If -we ask whether a;" — 1 is divisible by x — 1, it is easy to
make the actual division for any particular value oi n, as n = 2
or n = 3. But if x" — 1 is shown divisible by a; — 1, we are no
wiser than before concerning the divisibility of x* — 1. Suppose,
however, we can prove that the divisibility for n=m + l follows
from that for n = m, whatever value m may have. Then since
we have established the fact by direct division for w = 3, we may
be assured of the divisibility for ?i = 4, then for n = B, and so on.
Now a;'"+' — l=a;(a;'" — l) + (a; — 1)
is identically true. If a; — 1 is a factor of a;"* — 1 for a given value
of m, it is a factor of the right-hand member and consequently a
factor of the left-hand member (§ 69), which was to be proved.
Thus the divisibility of x" — 1 by a; — 1 is established for any
integral value of n greater than the one for which the division
has actually been carried out.
To complete^ the proof of a theorem by mathematical induction
we must make two distinct steps.
Mrst. Establish the theorem, for some particular case or cases,
preferably for n = 1 and n = 2.
Second. Show that the theorem for n =m + 1 follows from
its assumed validity for n = m.
Example. Prove that the sum of the cubes of the integers
from 1 to wis \]i{n{n + l)-]\\
To prove that 1» -i- 2' -t- 3' -(- ■ • • -|- w" = | ^ [»i (»H- 1)] j^.
125
126 QUADRATICS AND BEYOND
First. This theorem is true for n = l.
For l« = l = |^[l(2)]p = l'' = l.
The theorem is also true for n = 2.
For l« + 2» = 9= |^[2(2 + l)]P = (i-6)''=3^ = 9.
Second. Assume the theorem for n = m,*
1» + 2» + . . . + M« = \^[m{m + l)]p.
Add (m + 1)' to both sides of the equation,
18 + 2» + . . ■ + m» + (m + 1)« = J^[m(m + 1)] p + (m + If
= {_{^my + m + l-]{m + lf
=( --+7+^ )(^+i)^
= [*("» + !)('» + 2)?,
which is the form desired, i.e. m + 1 replaces m in the formula.
EXERCISES
Prove by mathematical induction that
1. 1+ 3+5 + .-.+(2n-l) =n2.
2. 2 + 22 + 23 + h- 2" = 2(2'> - 1).
3. 3 + 6 + 9H +3ri = |?i(n + 1).
4. 12 + 22 + 32 + • • • + n2 = ^n(n + 1) (2n + 1).
5. 18 + 28 + 33 + ■ ■ • + n8 = (1 + 2 + 3 + • . . + n)2.
6. 42 + 72 + 102 + . . . + {3n + 1)2 = in(6n2 + 15^ + 11).
7. a?" — 2/" is divisible by a; — y for any integral values of n.
8. s2m _ y2n jg divisible by x + y for any integral values of n.
9. 1.2 + 2-3 + 3-4 + 4.5H |-n-(n + l) = Jn(n + 1) (n + 2).
10. 1 ■ 1 + 2 ■ 32 + 3 • 52 + . ■ . + ji(2n - 1)2 = ^n(n + l)(6n2 _ 2n - 1).
11. 1.2 -3+2. 3.4+3.4 ■5 + --.+n(n+l)(n+2) = in(n+l)(n+2)(n+3).
12. (18 + 2» + 38 + • . . + to8) + 3(16 + 26 + 36 + . . . + „6)
= 4(1 + 2 + 3+ •■■ +n)8.
* This statement does not imply that we assume the validity of the formula for any
values for which it has not yet been established, but only for values of m not greater
than 2.
MATHEMATICAL INDUCTION 127
13. 1 ■ 3 + 3 . 32 + 5 . 3« + • . . + (2n - 1)3» = 3» + i(re - 1) + 3.
14 i„ + JL + ^ + -+ '
1-2 2.3 3-4 n-(ji + l) ji + 1
15. 12 + 32 + 62 + ... + (2n-l)^=="(^" + ^^'^"-^).
16. 2.6 + 3.6 + 4.7 + -.. + (n + l)(» + 4) = "<" + ^nn + 5)^
17. 2.4 + 4-6 + 6-8 + --- + 2n(27i + 2) = ^(2ji + 2)(2m + 4)
o
18. A pyramid of shot stands on ». triangular base having m shot on a
side. How many shot are in the pile ?
CHAPTEE XII
BINOMIAL THEOREM
126. Statement of the binomial theorem. When in prsTioiis
problems any power of a binomial has been required we have
obtained the result by direct multiplication. We can, however,
deduce a general law known as the binomial theorem, which gives
the form of development of (a + b)", where n is any positive integer
and a and b are any algebraical or arithmetical expressions. This
law is as follows :
(a + by =ar + '^ a"- '6 + '^-<>-^) a^^-^b^ + ■ • • + 6».
From this expression we deduce the following
EULE FOR THE DEVELOPMENT OF (a + &)".
The first term is a".
The second term is — a"~^6.
To obtain any term from the preceding term, decrease the
exponent of a in the preceding term hy 1 and increase the
exponent of b hy 1 for the new exponents. Multiply the coefficient
of the preceding term hy the exponent of a, and divide it hy the
exponent of h increased hy 1 for the new coefficient.
Remark. In practice it is usually more convenient first to write down all the
terms with their proper exponents, and then form the successive coefficients.
EXERCISES
Verify by multiplication the rule given for the following :
1. {a + 6)8. 2. (x - yy.
3. (2a + 36)*. 4. (\^+v^)'.
5. (2a-6)«. 6. {x--Jy)\
1. (3n-2 6)». 8. (a-^x + h-^y)*:.
128
BINOMIAL THEOKEM 129
127. Proof of the binomial theorem. We have aheady stated
(a + 5)" = a" + ^ a° -^6 + ''^^ ~ ^^ a'-'b" + ■■■ (1)
and have seen that it is justified for every particular case that
we have tested. By complete induction we now prove this
theorem when m is a positive integer.
First. Let n = 2.
That is, (a + by=a'' + 2ab + b\
This expression evidently obeys the law as stated in (1).
Second. Assume the theorem for ra = m.
That is, (a + S)™ = a" + ^ a^-^J + '^^'^ 7 ^^ a^'^b^ + • ■ ■. (2)
\ ^ 1 ■ 2i
Multiply both members by a. + 5,
(a + &)"•+' = »"■+! + ja'"b + j oT'-W
= «■"+' +(m + l)a'"b + ^™ "t ''•''"' a'-'-H^ + ■■■.
This expression is identical with (2) except that (m + 1)
replaces m. Hence the theorem is established so far as the
first three terms are concerned.
128. General term. Though we have stated the binomial
theorem for a general value of n, we have only established the
exact form of the first three terms.
Let (a + 6)» = a» + Cja"-'* + Cja"- W -\ .
We note that the sum of the exponents of a and S is n in any
term of the development of (as + J)". Also the exponent of 6 in
the (r + l)st term is r.
We have already seen that
n
n(n-~V)
"'-I'^^y 1.2
and that the first three terms are
1 1 • ^
130 QUADRATICS AND BEYOND
respectively. This indicates that the coefficient of the next term
■will be — ^^ — -too ^^^ ^^ general that the coefficient of the
(r + l)st term has the form
^ n(n-l)...(n-r + l)
l-2---r ^ ''
which is in fact the form that our rule (§ 126) would afford for
any particular value of r.
This affords the following
EuLE. The (r + l)st term of (a +1)" is
n{n-l)---(n-r + l) ^„_ , j,
1-2-r
The form of the coefficient may he easily remembered since the
denominator consists of the product of the integers from 1 to r,
while the numerator contains an equal number of factors consist,
ing of descending integers beginning with n.
For any particular values of n and r we could easily verify the
rule by direct multiplication. For the rigorous proof see p. 178.
EXERCISES
Develop by the binomial theorem :
1.
Solution :
/^Y+ ? /^Y/_^Y+ ^/^Y/^-2^Y+ ^iLi (^(^:^
6.5-4-3 /jlY/- '^+ 6-5-4-3.2 /_a_Y/'_:^°
1.2.3-4\Vi/V aV 1.2.3-4.5\^A a^ I
i • 6 \ a V
+ ■
6.S-4-3-2-
1.2-3-4.5-
_?!_6— 4- — - — + 15^-^ —
2.(1-10)". 3. (Va + ^)^
BINOMIAL THEOREM 131
6 (^-^Y 7 (^--X
' \Sv 2x/' ' \S xV'
\10 6/ \23/
10. (i+va)'-(i-vs)'. 11. (vi+v^y+wi-viy.
(2 X
— -
Solution : 71 = 10, r + 1 = 8.
The (r + l)st term of (a + 6)» is (§ 128)
n(n — 1) ■ ■ ■ {n — r + 1) ^
In this case we get
2x) '
1-2. ■ r
10.9.8-7-6.6.4 (^Y(W
1-2.3-4.5.6.7 [sy) \2x/
= i2o.?!5!.!V = i2o.?V
38j/8 2''a;' 2*x*
_ 120 -81 y* _1215y*
13. rind the 7th term of
14. Find the 6th term of
16 X* 2 a^
18
\2y x) '
15. Find the 8th term of ( — - ^ ) .
\ y X /
16. Find the 6th term of (2aVb — Y'-
V 2aVbJ
17. Find the 7th term of (~ - ^T.
18. Find correct to three decimal places (.9)'.
Solution: (.9)' = (1 -.1)8
= (i)« - 1 (!)'(• 1) + ^(i)M-i)^ - 1^ (i)M-i)"'
8^7^6^ _
1-2. 3-4^ / V ' -r
= 1 - 8 • 0.1 + 28 • 0.01 - 56 • 0.001 + 70 ■ 0.0001
= 1 + 0.28 + 0.0070 - 0.8 - 0.056
= 1.2870 - 0.856 = .431.
In this exercise any terms beyond those taken would not affect the first
three places in the result.
132 QUADRATICS AND BEYOND
Compute the following correct to three places :
19. (l.l)i». 20. (2.9)8. 21. (.98)". 22. (1.01)6.
23. (f|)8. 24. (|J)io. 25. (98)». 26. (203)6.
27. In what term of (a + by does a term involving a^* occur?
28. For what kind of exponent may a and 6 enter the same term with
equal exponents ?
29. For what kind of exponent is the number of terms in the binomial
development even ?
30. Find the first three and the last three terms in the development of
CHAPTEE XIII
ARITHMETICAL PROGRESSION
129. Definitions. A series of munbers such that each numbei
minus the preceding one always giyes the same positive or
negative number is called an arithmetical series or arithmetical
progression (denoted by A.P.).
The constant difference between any term and the preceding
term of an A.P. is called the common difference.
The series 4, 7, 10, 13, • • • is an A.P. with the common difference 3. The series
8, 6J, 5, 3i, ■■■ is an A.P. with the common difierence — i. The series 4, 6, 7, 9,
10, • • ■ is not an A.P.
EXERCISES
Determine whether the following series are in A.P. If so, find the common
difference.
1. 6, 3ii 1|, .... 2. 27, 22i, 18, ....
3. 6, 4J, 3, li, .... 4. 5, -2, -8, ....
5. Vi, -V^, 3VJ. •■•• 6. 8, 5J, 3|, If, ....
_ 1 2 4 V2-1 V2 1
7. ~-?=' ~;=' ~s' . 8, ■
V2 V2 -v^ ■ 2 ' 2 ' 2(V2-1)'
V3 3V3 + 2 -39
9. 3, -}, -3f, -6f, .... 10.
6 6 (V3 - 4)
130. The nth term. The terms of an A.P. in which a is the
first term and d the common difference are as follows :
a, a + d, a + 2d, a + Sd, ■ ■ ■. (1)
The multiple of d is seen to be 1 in the second term, 2 in the
third term, and in fact always one less than the number of the
term. If we call I the wth term, we have
1 = a +(n — V)d.
133
134 QUADRATICS AND BEYOND
We may also write the series ia which I is the «th term as
follows :
a, a -\- d, a + 2d, •■-, I — 2d, I — d, I.
131. The sum of the series. We may obtain a formula for
computing the sum of the first n terms of an A.P. by the following
Theorem. The sum s of the first n terms of the series
a, a + d, •••, I — d, I is
s = l{a + l).
By definition,
s = a +(a + d)-\-{a + 2d)-\ +{l - 2d) + {l- d)+ I. (1)
Inverting the order of the terms of the right-hand member,
s = l + {l-d) + Q-2d)-{ \-{a + 2d) + {a + d)+a. (2)
Adding (1) smd (2) term by term,
2s = (l + a)+{l + a) + {l + a)+--- + {l-\:a) + {l + a) + (l + a)
Thus s = ^ (a + Z).
132. Arithmetical means. The terms of an A.P. between a
given term and a subsequent term are called arithmetical means
between those terms. By the arithmetical mean of two numbers
is meant the number which is the second term of an arithmetical
series of which they are the first and third terms. Thus the
arithmetical mean of two numbers a and b is — r — > since the
numbers a, — ^r—> h are in A.P. with the common difference — — ■
The two formulas
l = a ■\-{n — V)d, (I)
s = J(« + ?) (II)
contain the elements a, I, s, n, d. Evidently when any three are
known the remaining two may be found by solving the two equa-
tions (I) and (II).
ARITHMETICAL PROGRESSION 135
EXERCISES
1. Find tlje 16th term and the sum of the series 4, 2, 0, — 2, ■ • •.
Solution: n = 16, a = 4, d = 2 - 4 = - 2.
l = a + {n-l)d = i + 15(- 2) = - 26,
s = |(a + J) = y(4-26) = -176.
2. J = 42, a = — 3, d = 3. Find n and s.
3. a = - 4, m = 8, s = 64. Find d and I.
4. d = — J, n = 6, J = 21. Find s and a.
5. d = — ^, K = 10, s = 65. Find a and !.
6. s = 16J, J = 4, a = - 3. Find d and n.
7. J = 22, s = 243, »i = 13. Find a and d.
8. s = - 15, J = - 2, d = 2. Find n and a.
9. d = 4i, o = - 16, s = 140. Find n and i.
10. Insert 8 arithmetical means between 4 and 28.
11. Find expressions for n and s in terms of a, I, and d.
12. Find expressions for I and a in terms of s, n, and d.
13. Find expressions for a and s in terms of d, !, and n.
14. Find expressions for d and n in terms of s, a, and Z.
15. Find the 13th term and the sum of the series
V2-1 V2 1
2 ' 2 '2(V2-1)'"'"
16. Find the 10th term and the sum of the series
V3 3V3 + 2 V3 2
T' 6 'T'+3'"-
17. Insert 4 arithmetical means between —- and
V2 2
18. Insert 5 arithmetical means between -» /- and — - — •
19. Insert 3 arithmetical means between — and
2 6
20. Find the 21st term and the sum of the series —-, V2, , ■ • •.
V2 2
21. Find the 10th term and the sum of the series — =, — -, ■ — -- , • • • .
V3 V6 V3
136 QUADRATICS AKD BEYOND
22. rind expressions for d and a in terms of s, I, and n.
23. Find expressions for d and Z in terms of a, n, and ».
24. Find the 8th term and the sum of the series z, ix, Tx, ■••.
25. Find the 9th term and the sum of the series 8, 9|, lOf, • • •.
26. Find the 12th term and the sum of the series 8, 7^, 6J, • • • .
27. Find the 8th term and the sum of the series — 8, — 4, 0, • • •.
28. Find the 12th term and the sum of the series 27, 22 J, 18, • • • .
29. Find the 20th term and the sum of the series 1, — 2|, — 6, ■ • •.
30. Find the 11th term and the sum of the series 5, — 3, — 11, ■ ■ •.
31. Find the 9th term and the sum of the series x — y, x, x + y, ■ ■ ■.
32. Insert n — 2 arithmetical means between a and I. Write the first three.
Kemabk. Often an exercise may be solved more simply if instead of assum-
ing the series x, x-\-y, x + 2y, ■■ ■ yfe assume x — y, x,x + y when three terms
are required, orx — 2y,x — y,x,x + y,x + 2y when five terms are required, or
X — Sy, X — y, x + y,x + Sy when four terms are required.
33. The sum of the first three terms of an A. P. is 15. The sum of their
squares is 83. Find the sum of the series to ten terms.
34. Find expressions for n and a in terms of s, I, and d. For what real
values of s, I, and d does a series with real terms not exist ?
35. In an A.P. where a is the first term and s is the sum of the first
n terms, find the expression for the sum of the first m terms.
36. Find expressions for n and I in terms of o, s, and d. For what real
values of a, s, and d does a series with real terms not exist ?
37. If each term of the series (1), § 180, is multiplied by to, is the new
series in A.P., and if so, what are the elements of the new series ?
38. If each term of the series (1) in § 130 is increased by 6, is the new
series in A.P., and if so, what are the elements of the new series ?
39. The difference between the third and sixth terms of an A.P. is 12.
The sum of the first 10 terms is 45. Find the elements of the series.
40. Find the 10th term of an A.P. whose first and sixteenth terms are 3
and 48. Find also the sum of those eight terms of the series the last of
which is 60.
41. Two A.P.'s have the same common difference, and their first terms
are 2 and 4 respectively. The sum of the first seven terms of one is to the
sum of the first seven terms of the other as 4 is to 5. Find the elements of
both series.
42. The three digits of a number are in A.P. The number itself divided
by the sum of the digits is 48. The number formed by the same digits in
reverse order is 396 less than the original number. What is the number?
CHAPTEE XIV
GEOMETRICAL PROGRESSION
133. Definitions. A series of numbers sucli that the quotient
of any term of the series by the preceding term is always the
same is called a geometrical progression (denoted by G.P.).
The constant quotient of any term by the preceding term of a
G.P. is called the ratio.
The G.P. series 4, 8, 16, ■ • • has the ratio 2. The G.P. aeries 8, 4 Vi, 4, ■ • ■ has
the ratio .
^ EXERCISES
Determine which of the following series are in G.P. ,and find the ratio.
1. 4, 2, 1, ....
3. 8, -2, .5, ....
5. Vf, 1, Vf, •••.
. 7.-^, V2,2, ....
V2
11. —-J^ — =, 5-2V6,3V3-V2,..
V3-V2
134. The jith term. The terms of .a. G-.P. in which a is the
first term and r the ratio are as follows :
a, ar, ar^, ar^, ■■-.
The power of r in the second term is 1, in the third term is
2, and in fact is always one less than the number of the term.
If we call I the wth term, we have the following expression for
the wth term :
l=ar^-\ (I)
137
2.
4,8,
, 16, ■■-.
4.
8, -
4,-2, ....
6.
6, -
21, 73i, ....
8.
1
vi'
-•A
10.
V3
8 '
[W Vs
V32'X'"
12.
V2.
-1, 1, ^^^-l,
138 QUADRATICS AND BEYOND
135. The sum of the series. We obtain a formula for finding
the sum of the first n terms of a G.P. by the following
Theorem. The sum s of the first n terms of the geometrical
progression a, ar, ar^, ...is
a — rl
' = T^r
By definition, s = a-\- ar -\- wr^ -\ + (ir^~^
= aCY -k- r -\- r^ -\ \- r^-"^)
"(^)
a — rl
1-r'
by (III), p. 16
by (I), p. 137
136. Geometrical means. The n — 2 terms between the first
and the rath term of a G.P. are called the geometrical means
between those terms.
If one geometrical mean is inserted between two numbers, it
is called the geometrical mean of those numbers. Thus the
geometrical mean between a and h is Va6.
The two fundamental formulas
« = «»•''- 1, (P)
a(l — r'^) a — rl ,,.-..
s = -\ i = (II)
1 — r 1 — r ^'
contain the five elements a, I, r, n, s, any two of which may be
found if the remaining three are given.
EXERCISES
1. Find the 7th term and the sum of the G.P. 1, 4, 16, ■ ■ •.
Solution :
Substituting in (I),
Substituting in (11), s
a ■■
= 1, Ji =
= 7,
r = 4.
1 =
ar»-i =
= 1-
46 = 4096,
S z
a
-rl
— r
1
-4
■ 4096
= 1«^«^-5461
GEOMETRICAL PKOGRESSIOjST 139
2. Insert 2 geometrical means between 4 and 32.
3. Insert 4 geometrical means between 32 and 1.
4. Insert 3 geometrical means between 3 and Jf.
5. Insert 4 geometrical means between a^ and b^.
6. Insert 4 geometrical means between 1 and 9V3.
7. Insert 3 geometrical means between if- and 73|.
8. What is the geometrical mean between 3 and 27 ?
9. Insert 3 geometrical means between VS and V^i.
10. Insert 4 geometrical means between a and a" Va6^
11. Insert 3 geometrical means between — f and — 2J.
12. What is tlie geometrical mean between — 2 and — | ?
13. rind the 7th term and the sum of the series 1, 3, 9, • • • .
14. Find the 6th term and the sum of the series 2, 4, 8, ■ • ■.
15. What is the geometrical mean between Vein and Val^?
16. Find the 7th term and the sum of the series 8, 2, . 5, • • ■ .
17. Find the 8th term and the sum of the series ^\, iVi ii ' ' ' •
18. Find the 7th term and the sum of the series Vi, 2, 2-v/2, • ■ •.
19. Find the 7th term and the sum of the series v2, v^, V4, • • • .
20. Find the 10th term and the sum of the series j^, ^ Jj, ^^j, • • • .
21. Find the 5th term and the sum of the series V2 — 1, 1, 1 + V2, • • ■.
22. The first and sixth terms of a G.P. are 1 and 243. Find the interme-
diate terms.
23. Find the 5th term and the sum of the series
, 5-2V6, 9V3-IIV2, ■...
24. Insert 3 geometrical means between — = and -.
V3 9
25. What is the geometrical mean between and - '
x+y x-y
26. Find the 6th term and the sum of the series — =, — V2, , ■
V2 V2
27. Find the 6th term and the sum of the series .» (- , 1, — 1 • ■ •.
\3 V2
V2
28. Find the 5th term and the sum of the series ■s/2, — 1, — , . . •
i9. Find the 6th term and the sum of the series , .» / — , — , •
8 \32 4
140 QUADKATICS AND BEYOND
30. The geometrical mean of two numbers is 4 and their sum is 10. Find
the numbers.
31. The fourth term of a G.P. Is 192, the seventh term is 12,288. Find
the first term and the ratio.
32. If the same number be added to or subtracted from each term of a
G.P., is the resulting series geometrical?
33. The product of the first and last of four numbers in G.P. is 64.
Their quotient is also 64. Find the numbers.
34. The product of four numbers in G.P. is 81. The sum of the second
and third terms is J. Find the numbers.
35. If every term of a G.P. be multiplied by the same number m, is the
resulting series a G.P.? If so, what are the elements?
36. The sum of three numbers in G.P. is 42. The difference between the
squares of the first and the second is 60. What are the numbers ?
37. The difference between two numbers is 48. The arithmetical mean
exceeds the geometrical mean by 18. Find the numbers.
38. Four numbers are in G.P. The difference between the first and the
second is 4, the difference between the third and the fourth is 36. Find the
numbers.
39. A ball falling from a height of 60 feet rebounds after each fall one
third of the last descent. What distance has it passed over when it strikes
the ground for the eighth time ?
40. The difference between the first and the last of three terms in G.P.
is four times the difference between the first and second terms. The sum of
the numbers is 208. Find the numbers.
41. An invalid on a certain day was able to take a single step of 18
inches. If he was each day to walk twice as far as on the preceding day,
how long before he can take a five-mile walk ?
42. The difference between the first and the last of four numbers in G.P.
is thirteen times the difference between the second and third terms. The
product of the second and third terms is 3. Find the numbers.
137. Infinite series. Wten the number of terms of a G.P. is
unlimited it is called an infinite geometrical series.
In the series a, ar, ar^, ■■■, when r > 1, evidently each term is
larger than the preceding term. The series is then called increas-
ing. "When r < 1, each term is smaller than the preceding term
and the series is called decreasing.
„ . a (1 — r") a ar"
Now m any case s = —^ ' = z z
■' 1 — r 1 — rl — r
GEOMETKICAL PROGRESSION 141
Wien r > 1, evidently r" becomes very large for large values
of 11. For this case, then, the sum of the first n terms becomes very
large for large values of n. In fact we can take enough terms
so that s will exceed any number we may choose. If, however,
r < 1, as w increases in value r" becomes smaller and smaller. In
fact we can choose n large enough so that r" is as small as we wish,
or as we say, approaches as a limit. But since r" may be made
as small as we wish, ar" also approaches as a limit, and conse-
quently :j approaches as a limit. Thus when r < 1 the
value of the sum of the first n terms approaches as n
becomes very great. This we express in other words by asserting
that the sum of the infinite series
a + ar + ar' + • • •, when r < 1,
" 1 - r
EXERCISES
Find the sum of the following infinite series.
1. 6 + 3 + I + ■ • ..
Solution : a = 6, )• = ^.
2. l+i + i + ---. 3. 64 + S + 1+---.
4. i + i + |+---. 5. i + T^ + A + ---
6. | + f + -V + ---- 7. 2 + .5 + .125 + ..-.
-i/2
». V2 + l + ^ + ---. 9. (■^/2 + l) + l + (^^-l) + ....
10. How large a value of n must one take so that the sum of the first
n terms of the following series differs from the sum to infinity by not more
than .001 ?
(a) 8 + 4 + 2 + ..-.
Solution : a = 8, r = J.
_ a ar^ _ or"
1 — r 1 — r 1 — r
Soo — 8= :;
\ — r
142 QUADRATICS AND BEYOND
We must find for what value of n the expression ■ is less than .001.
8(|)''_8-2_16
1 - J ~ 2" ~ 2»'
By trial we see that if n = 14 the value of — is , which is less than
.001. 2« 1024
(b) 27 + 3 + i + .... (c) 4 + i + jij+....
(d) l + TV + 5k + ---- (e) 64 + 16 + 4 + ....
(f) 100 + 20 + 4 + . • .. (g) 60 + 20 + 6f + . • ..
11. What is the value of the following recurring decimal fractions?
(a) .212121....
Solution : This decimal may be written in the form
Here
(b) .333....
(e) .343343.
21+ '' + '"■ +....
100 (100)2 (100)3
21 1
a = — ■* r = — •
100 100
a .21 .21 7
'" 1-j- l-.Ol .99 33
(c) .717171. ••.
(d) .801801.-
(f) 1.43131....
(g) 2.61414..
ADVANCED ALGEBRA
CHAPTEE XV
PERMUTATIONS AND COMBINATIONS
138. Introduction. Before dealing directly witii the subject of
the chapter we must answer the qtiestion, In how many distinct
ways may two successive acts be performed if the first may be
performed in -p ways and the second may be performed in q ways ?
Suppose for example that I can leave a certain house by any
one of four doors, and can enter another house by any one of five
doors, in how many ways can I pass from one house to the other ?
If I leave the first house by a certain door, I have the choice of
all five doors by which to enter the second house. Since, how-
ever, I might have left the first by any one of its four doors,
there are 4 • 5 = 20 ways in which I may pass from one house
to the other. This leads to the
Theorem. If a certain act may he performed in p ways, and
if after this act is performed a second act may he performed in q
ways, then the total number of ways in which the two acts may
he performed is p ■ q.
With each of the p possible ways of performing the first act
correspond q ways of performing the second act. Thus with all
the p possible ways of performing the first act must correspond
q times as many ways of performing the second act. That is, the
two acts may be performed in ^ • g' ways.
It is of course assumed in this theorem that the performance
of the second act is entirely independent of the way in which the
first act is performed.
143
144 ADVANCED ALGEBRA
EXERCISES
1. I have four coats and five hats. How many difEerent combinations of
coat and hat can I wear ?
Solution : The first act consists in putting on one of my* coats, which may
be done in four ways; the second act consists in putting on one of my
hats, which may be done in five ways. Thus I have 4 ■ 5 = 20 difEerent
combinations of coat and hat.
2. In how many ways may the two children of a family be assigned to
five rooms if they each occupy a separate room ?
3. A gentleman has four coats, six vests, and eight pairs of trousers. In
how many difEerent ways can he dress ?
4. I can sail across a lake in any one of four sailboats and row back
in any one of fifteen rowboats. In how many ways can I make the trip ?
5. Two men wish to stop at a town where there are six hotels but do not
wish quarters at the same hotel. In how many ways may they select hotels ?
6. A man is to sail for England on a steamship line that runs ten'boats
on the route, and return on a line that runs only six. In how many difEerent
ways can he make the trip ?
7. In walking from A to B one may follow any one of three roads; in
going on from B to C one has a choice of five roads. In how many different
ways can one walk from A to C ?
139. Permutations. Each difEerent arrangement either of all
or of a part of a number of things is called, a permutation.
Thus the digits 1, 2 have two possible permutations, taken both
at a time, namely, 12 and 21.
The digits 1, 2, 3 have six different permutations when two are
taken at a time, namely, 12, 13, 21, 23, 31, 32. For if we take 1
for the iirst place, we have a choice of 2 and 3 for the second
place, and we get 12 and 13. If 2 is in the first place, we get 21
and 23. Similarly, we get 31 and 32. In this process it is noted
that we can fiU the first of the two places in any one of three
ways ; the second place can be filled in each case in only two ways.
Thus by the Theorem, § 138, we should expect 3-2 = 6 permuta-
tions of three things taken two at a time. We observe that this
product 3 • 2 has as its first factor 3, which is the total number of
things considered. The number of factors is equal to the number
of digits taken at a time, i.e. two. This leads to the general
PERMUTATIONS AND COMBINATIONS 145
Theoeem. The number of permutations of n objects taken r
at a time is i\ , , -i\ /t\
n{n — l)---{n — r + I). (I)
This is symbolized by P„^ ^.
This formula is easily remembered if one observes that the iirst
factor is n, the total number of objects considered, and that the
number of factors is r, the number of objects taken at a time.
Thus P7,3 = 7-6-5.
We prove this theorem by complete induction.
First, let r = 1. There are evidently only n different arrange-
ment of n objects, taking one object at a time, namely (assuming
our objects to be the first n integers),
1, 2, 3, .., n.
Let us take two objects at a time, i.e. let »• = 2. Since there
are n objects, we have n ways of filling the first of the two places.
When that is filled there are n — 1 objects left, and any one may
be used to fill the second place. Thus, by the Theorem, § 138,
there ' are for r = 2 , ..
n(n — 1)
different permutations.
Second, assume the form (I) for r = m,
Pn.m = '^(n-l)---in-m + l). (1)
We can fill the first m places in P^ „ different ways since there
are that number of permutations of n things taken m at a time.
This constitutes the first act (§ 138). The second act consists in
filling the m + 1st place, which may be done in n — m ways by
using any of the remaining n — m objects. Thus the number of
permutations of n things taken m + 1 at a time is
■Pn, m+i = -Pn, m • ('I — ™) = ''■ (™ — 1) (™ — 2) ■ • ■ (ra — m + 1) (n — m),
which is the form that (1) assumes on replacing mhj m + 1.
CoEOLLAEY. The number of permutations of n things taken all
atatimeis p^^ = n{n- l)---2-l = n!* (2)
Taking n ^r m (I), we get (2).
* w / is tlie symbol for 1 • 2 ■ 3 • 4 • ■ ■ (w - 1) ?i, and is read/oc^oriaZ n.
146 ADVANCED ALGEBRA
EXERCISES
1. How many permutations may be formed from 8 letters taken four at a
time?
Solution : n = 8, r = 4, n — r + l=5,
Ps,4 = 8-7-6-5 = 1680.
2. In how many different orders may 6 boys stand in a row ?
3. How many different numbers less than 1000 can be formed from the
digits 1, 2, 3, 4, 5 without repetition ?
4. How many arrangements of the letters of the alphabet can be made
taking three at a time ?
5. How many numbers between 100 and 10,000 can be formed from the
digits 1, 2, 3, 4, 5, 6 without repetition ?
6. How many different permutations can be made of the letters in the
word compute taking four at a time ?
7. In a certain class there are 4 boys and 5 girls. In how many orders may
they sit provided all the boys sit on one bench and all the girls on another ?
Hint. Use Corollary § 139, and then Theorem, § 138.
8. I have 6 books with red binding and 3 with brown. In how many ways
may I arrange them on a shelf so that all the books of one color are together ?
140. Combinations. Any group of things that is independent of
the order of the constituents of the group is called a combination.
The committee of men Jones, Smith, and Jackson is the same
as the committee Jackson, Jones, and Smith. The sound made by-
striking simultaneously the keys EGrC of a piano is the same as
the sound made by striking CGE. In general a question involv-
ing the number of groups of objects that may be formed where
the character of any group is unaltered by any change of order
among its constituent parts is a question in combinations.
Suppose for example that we ask how many committees of three
men can be selected from six men. If the men are called A, B, C,
D, E, F, there are, by § 139, 6 • 5 ■ 4 = 120 different arrangements
or permutations of the six men in groups of three. But the permu-
tations A, B, C ; A, C, B ; B, A, C, etc. (3 ! = 6 in all for the men
A, B, and C), are all distinct, while evidently the six committees
consisting of A, B, and C are identical. This is true for every
distinct set of three men that we could select; that is, for the
PERMUTATIONS AND COMBINATIONS 147
six different permutations of any three men there is only one
distinct committee. Hence the number of committees is one sixth
P
the total number of permutations, or -^•
o I
This leads to the general
Theorem. The number of combinations of n things taken r at
a time is , -.^ , . -.^
n{n — I)- ■■ (n — r + 1)
71
This is symbolized by C„^ ^.
The number of permutations of n things taken r at a time is
P„,, = «.(™-l)---(?:-r + l).
In every- group of r things which form a single combination
there are (Cor., p. 145) r ! permutations. Thus there are r ! times
as many permutations as combinations. That is,
r ^--r n(n~l)-.-(n-r+l)
This formula is easily remembered if one observes that there
is the same number of factors in the numerator as in the denomi-
nator. Thus
1098
^10,3- 1.2.3
COEOLL AEY. C„^ ^ = C„_ „ _ ^ .
Multiplying numerator and denominator of (I) by (n — r)l,
__ n(n — !)■ ■ ■ (n — r + 1) (n — r) ■ ■ ■ 2 -1
»■'•"" rl(n-ry.
r'.n — rl
_ n(n-l)---(r + l)
(n — ry.
_ n(n — !)■ ■ -[n — (n ^ r) + T\
(n — r)\
= ^n, n — r-
This corollary saves computation in some cases. For instance, if we wish to
19-18
compute Ci9, 17, it is more convenient to write Cia, 17 = C19, 2 = — — — = 171 than the
expression for Cm, 17.
148 ADVANCED ALGEBRA
EXERCISES
1. How many committees of 5 men can be selected from a body of 10
men three of whom can serve as chairman but can serve in no other capacity ?
Solution : There are 7 men who may fill 4 places on the committee.
'■ 1.2.3.4
There are 3 men to select from for the remaining place of chairman,
and the selection may be made in 3 ways. Thus the committee can be
made up in 3 • 85 = 105 ways.
2. How many distinct crews of 8 men may be selected from a squad of
14 men ?
3. How many distinct triangles can be drawn having their vertices in
10 given points no three of which are in a straight line ?
4. How many distinct sounds may be produced on 9 keys of a piano by
striking 4 at a time ?
5. In how many ways can a crew of 8 men and a hockey team of 6 men
be made up from 20 men ?
6. In how many ways may the product a-b-c-d-e-f\)6 broken up
into factors each of which contains two letters?
7. If 8 points lie in a plane but no three in a straight line, how many
straight lines can be drawn joining them in pairs ?
8. How many straight lines can be drawn through n points taken in
pairs no three of which are in the same straight line ?
9. Seven boys are walking and approach a fork in the road. They
agree that 4 shall turn to the right and the remainder turn to the left. In
how many ways could they break up ?
Solution : The number of groups of 4 boys that can be formed from the
O7, 4 = — = o5.
4!
Por each group of 4 boys there remains only a single group of 3 boys.
Thus the total number of ways in which the party can divide up is
precisely 35.
10. If there are 12 points in space but no four in the same plane, how
many distinct planes can be determined by the points ?
Hint. Three points determine a plane.
11. Bight gentlemen meet at a party and each vrishes to shake hands
with all the rest. How many hand shakes are exchanged ?
PERMUTATIONS AND COMBINATIONS 149
12. In how many ways can a baseball team of 9 men be selected from
14 men only two of whom can pitch but can play in no other position ?
13. How many baseball teams can be selected from 15 men only four of
whom can pitch or catch, provided these four can play in either of the two
positions but cannot play elsewhere ?
14. Two dormitories, one having 3 doors, the other having 5 doors, stand
facing each other. A path runs from each door of one to every door of the
other. How many paths are there ?
15. Show that the number of ways in which p + g things may be divided
(n -u qY
into groups of p and a things respectively is — — i^.
plql
16. Out of 8 consonants and 3 vowels how many words can be formed
each containing 3 consonants and 2 vowels ?
17. A boat's crew consists of 8 men, three of whom can row only on one
side and two only on the other. In how many ways can the crew be arranged ?
18. A pack of cards contains 52 distinct cards. In how many different
ways can it be divided into 4 hands of 13 cards each ?
19. Five points lie in - plane, but no three in any other plane. How
many tetrahedrons can be formed with these points taken with two points
not in the plane ?
141. Circular permutatioas. By circular permutations we
mean tlie various arrangements of a group of things around a
circle.
Theoeem. The number of orders in which n things may be
arranged in a circle is (n—T) !.
Suppose A is at the point at which we begin to arrange the
digits 1, 2, 3, ■ • ■, n. Suppose we start our arrangement of digits
at A with a given digit a. We have then
virtually re — 1 places to fill by the remaining
ra — 1 digits. Thus we get (ra — 1) ! (p. 146)
permutations of the n digits keeping a fixed.
But suppose we start our arrangement, that is,
fill the place at A with any other digit, as 6,
and the remaining places in any order what-
ever. If we now go around the circle till we
come to the digit a, the succession of digits from that point
around the circle to a again must be one of the (w — 1) ! orders
150 ADVANCED ALGEBRA
■which we obtained when we took a as the initial figure. Thus
the only distinct orders in which the n digits can be arranged
on a circle are the {n — 1) ! permutations we obtained by fiUing
the first place with a.
EXERCISES
1. In how many orders can 6 men sit around a circular table ?
Solution :
n = 6, n-l = 5, (n - 1)! = 5 ! = 120.
2. In how many ways can 8 men sit around a circular table ?
3. In how many ways may the letters of live be arranged on a circle ?
4. In how many ways may the letters of permutation be arranged on a
circle ?
5. In how many ways can 4 men and 4 ladies sit around a table so that
a lady is always between two men ?
6. In how many ways may 4 men and their wives be seated around a table
so that no man sits next his wife but the men and the women sit alternately ?
7. In how many ways can six men and their wives be seated around a
table so that each man sits between his wife and another lady ?
8. In how many ways can 10 red flowers and 5 white ones be planted
around a circular plot so that two and only two red ones are adjacent ?
142. Theorem. The number of permutations of n things of
n'
which p are alike, taken all together, is —^•
If all the things were different, we should have n ! permutations.
But since p of the n things are alike, any rearrangement of those
p like things will not change the permutation. Por any fixed
arrangement of the n things there are p ! different arrangements
of the p like things. Thus — - of the n ! permutations are iden-
tical, and there are only — - distinct permutations of the n things
p of which are alike. '
Corollary. If of n things p are of one kind, q of another
kind, r of another, etc., then there are — ; — --^— — permutations
of the n things taken all at a time. -^ ' ^ '
PERMUTATIONS AND COMBINATIONS 151
EXERCISES
1. How many distinct arrangements of the letters of the word Cincinnati
are possible ?
Solution : There are in all 10 letters, of which 3 are i, 2 are c, and 3 are n.
Thus the number of arrangements is
10! _; r.2.g.^.6.^-7-8.9.10
3I3!2!~ X-^'-^X-^-^-X-?
= 2-6-7-8-9-10 = 50,400.
2. How many distinct arrangements of the letters of the word parallel
can be formed ?
3. How many signals can be made by hanging 15 flags on a staff if 2
flags are white, 3 black, 5 blue, and the rest red ?
4. How many signals can be made by the flags in exercise 3 if a white
one is at each extreme ?
5. How many signals can be made by the flags in exercise 3 if a red flag
is always at the top ?
6. Would 3 dots, 2 dashes, and 1 pause be enough telegraphic symbols
for the letters of the English alphabet, the numerals, and six punctuation
CHAPTEE XVI
COMPLEX NUMBERS
143. The imaginary unit. When we approached the solution
of quadratic equations (p. 52) we saw that the equation x" = 2
was not solvable if we were at liberty to use only rational num-
bers, but that we must introduce an entirely new kind of number,
defined as a sequence of rational niMnbers, if we wished to solve
this equation. The excuse for introducing such numbers was not
that we needed them as a means for more accurate measurement,
— the rational numbers are entirely adequate for all mechanical
purposes, — but that they are a mathematical necessity if we
propose to solve equations of the type given.
A similar situation demands the introduction of still other
numbers. If we seek the solution of
x' = -l, (1)
we observe that there is no rational number whose square is — 1.
Neither can we define V— 1 as a sequence of rational numbers
which approach it as a limit. We may write the symbol V— 1, but
its meaning must be somewhat remote from that of V2, for in
the latter case we have a process by which we can extract the
square root and get a number whose square is as nearly equal to
2 as we desire. This is not possible in the case of V^l. In fact
this symbol differs from 1 or any real number not merely in
degree but in kind. One cannot say V— 1 is greater or less than
a real number, any more than one can compare the magnitude of
a quart and an inch.
V— 1 is symbolized by i and is called the imaginary unit. The
term " imaginary " is perhaps too firmly established in mathe-
matical literature to warrant its discontinuance. It should be
kept in mind, however, that it is really no more and no less
152
COMPLEX NUMBERS 153
imaginary than the negative numbers or the irrational numbers
are. So far as we have yet gone it is merely that which satis-
fies eq[uation (1). "When, however, we have defined the various
operations on it and ascribed to it the various characteristic
properties of numbers we shall be justified in calling it a
number.
Just as we built up from the unit 1 a system of real numbers,
so we build up from V— 1 = i a system of imaginary numbers.
The fact that we cannot measure V— 1 on a rule should cause
no more confusion than our inability exactly to measure V2 on a
rule. Just as we were able to deal with irrational numbers as
readily as with integers when we had defined what we meant
by the four operations on them, so will the imaginaries become
indeed numbers with which we can work when we have defined
the corresponding operations on them.
144. Addition and subtraction of imaginary numbers. We
write
= i,
i + i = 2i,
i + i + i=3i,
i + i + ■ ■■ + i = ni. (I)
n terms
Also just as we pass from a rational to an irrational multiple
of unity by sequences, so we pass from a rational to a n irr ational
multiple of the imaginary unit. Thus we write a V— 1, or at,
where a represents any real number. Consistently with § 76 we
write
-tV-a'' = ±Va=-(-l) = ± Vo^- V^ = ±aV^.= ±ai. (II)
We speak of a positive or a negative imaginary according as
the radical sign is preceded by a positive or a negative sign.
We also define addition and subtraction of imaginaries as
"follows :
ai±bi = (a ± b) i, (III)
where a and b are any real numbers.
154 ADVANCED ALGEBRA
Assumption. The commutative and associative laws of multi-
plication and addition of real numbers,'^ 10, we assume to hold
for imaginary numbers.
145. Multiplication and division of imaginaries. We have
already virtually defined the multiplication of imaginaries by
real numbers by formula (I). Consistently with. § 76 we define
V^- V^= i i = i" =- 1.
Thus V— a ■ V— 6 = Va • y/b i i = Va6 •(—!) = — VaS.
The law of signs in multiplication may be expressed verbally
as follows :
The prodicet of imaginaries with like signs before the radical
is a negative real number. The product of imaginaries with
unlike signs is a positive real number.
For instance, - V^^ • V^l = - 2 • 3 • £= = 6.
We also note that
i^ = — 1, i' — — i, i* = 1, i^ = i, • ■ ■.
And, in general, i4» + A = j*^ fc = 0, 1, 2, 3.
We define division of imaginaries as follows :
/ / — 7 ~wa-i la
V- a -i- V- b = = A /-•
■Vb-i ^b
In operating with imaginary numbers, a number of the form
V— a should always be written in the form Vai before per-
forming the operation. This avoids temptation to the following
error * .^— — ____-^^^^^
V- a ■ V^ = ■V(- a) ■(-b) = Vab.
EXERCISES
Simplify the following :
1. V^l • V^^.
Solution: V^^- V^^ = V8-i- V2 ■ i = V2 • 8 • i^ = 4 . (_ 1) = _ 4.
2.1.
„ , . . 1 i' i' — i
Solution : — = — = = = — I.
i6 is (i4)3 I
COMPLEX NUMBEllS
155
3.
j".
6.
V-36.
9.
V— a;^".
12.
V2 V- 8.
15.
1
1R
V-a
4.
i2*.
5.
jU.
7.
V-64.
8.
11.
2 i • 3 i.
10.
V-3!c2ai
V-cc".
13.
V-2 V-6.
14.
V-3-.
16.
1
i
17.
V-4
V-2
19.
V-ft
20.
V-i*.
146. Complex numbers. The solution of the quadratic equation
with negative discriminant (p. 71) affords us an expression which
consists of a real number connected with an imaginary number
by a + or — sign. Such an expression is called a complex number.
It consists of two parts which are of different kinds, the real
part and the imaginary part. Thus 6 + ii means 6 I's + 4 i's.
Obviously, to any pair of real numbers (x, y) corresponds a complex
number x + iy, and conversely.
147. Graphical representation of complex numbers. "We have
represented all real numbers on a single straight line. When we
wished to represent two numbers simultaneously, we made use of
the plane, and assumed a one-to-one correspondence between the
points on the plane and the pairs of numbers (x, y). The general
complex number x + iy depends
on the values of the independent
real numbers x and y, and may
then properly be represented by
a point on a plane. We repre-
sent real numbers on the X axis,
imaginary numbers on the Y axis,
and the complex number x + iy
by the point (x, y) on the plane.
Thus the complex numbers 6 -(- i 3,
— 4 -I- i 4, 7 — 4 5, — 2 — * 4 are represented by points on the plane
as indicated in the figure.
148. Equality of complex numbers. We define the two com-
plex numbers a -\- ib and c + id to be equal when and only when
a = c and h = d.
y'
•.
-
4+
14
6+
i.3
T
2^
a
7-
i.")
156 ADVANCED ALGEBRA
SymtolicaUy a + ib=.c + id
when and only when a = c,b = d.
The definition seems reasonable, since 1 and i are different in
kind, and we should not expect any real multiple of one to cancel
any real multiple of the other.
Similarly, if we took not abstract expressions as 1 and i for
units but concrete objects as trees and streets, we should say that
a trees + h streets = c trees + d streets
when and only when a = c and h = d.
Principle. Wlien two numerical expressions involving imagi-
narieg are equal to each other, we may equate real parts and
imaginary parts separately.
The graphical interpretation ol the definition of equality is that equal complex
numbers are always represented by the same point on the plane.
From the definition given we see that a -\-ih = when and
only when a = S = 0.
Assumption. We assume that complex numbers obey the com-
mutative and associative laws and the distributive law given in
§ 10. We also assume the same rules for parentheses as given
in § 15.
This assumption enables us to define the fundamental opera-
tions on complex numbers.
149. Addition and subtraction. By applying the assumptions
just made we obtain the following symbolical expression for the
operations of addition and subtraction of any two complex num-
bers a + ib and g + id:
a + i5 ± (c + t(^ = a ± c + i(5 ± (^.
EuLE. To add (subtract) complex numbers, add (subtract) the
real and imaginary parts separately.
150. Graphical representation of addition. We now proceed
to give the graphical interpretation of the operations of addition
and subtraction.
COMPLEX NUMBEKS
157
,-C.-^D
E X
Theorem. The sum of two numbers A = a + ib and B = c + id
is represented by the fourth vertex of the parallelogram formed
on OA and OB as sides.
Let OA SB be a parallelogram. Draw
ES J. OE, AH _L ES, BD (= d) ± OE.
A AHS = A ODB since their sides are
parallel, and 0B = AS.
Thus DB = HS = d,
OD = AH = c.
Tarn ES = EH + HS = b + d,
OE = OF + FE = a + c,
and <S has coordinates (a -\- c, b + d) and represents the siira of
A and B, by § 149.
EXERCISES
1. The difierence A — B ot two numbers A = a + ib and B = c + id is
represented by the extremity D of the line CD drawn from the origin par-
allel to the diagonal BA of the parallelogram formed on OB and OA as
sides.
2. Represent graphically the following expressions.
(a) 1 + i. (b) - 4 - 2 i.
(c) 6 - i (d) - 8 + 4 i.
(6)2 + 41. (f) (1 + i) + (2 + i).
(g) (2-i)-(6-3i). (h) (l-j)-(l-2i).
(i) (2 + 4i) - (1 - 3i). 0") 4(1 + i)- 2(2 -Si).
(k) (6-2i) + (2 + 3i). (1) (6 + 3t) + (-l-6i).
151. Multiplication of complex numbers. The assumption of
§ 148 enables us to multiply complex numbers by the following
Etjle. To multiply the complex number a + ib by o + id, pro-
ceed as if they were real binomials, heeping in mind the laws for
multiplying imaginaries.
Thus a + ib
e + id
ac + icb + iad + (i)' bd = ac — bd + i(d> + ad).
158 ADVANCED ALGEBRA
152. Conjugate complex numbers. Complex numbers that differ
only in the sign, of their imaginary parts are called conjugate com-
plex numbers, or conjugate imaginaries.
Theorem. The sum avd the product of conjugate complex
numbers are real numbers.
Thus a + ib + a — ib = 2a,
(a + ib) (a - ib) = a^ + b\
153. Division of complex numbers. The quotient of two com-
plex numbers may now be expressed as a siagle complex number.
EuLE. To express the quotient — in the form x + iy,
rationalize the denominator, using as a rationalizing factor
the conjugate of the denominator.
a + ib a -\- ib c — id
Thus
■id -\- id — id
ac + bd — i (ad — be)
~ c' + d^
ac -\-bd .ad — be
(1)
c» + rf' C* 4- d^
We have now defined the fundamental operations on complex
numbers and shall make frequent use of them. If the question
remains in one's mind, "After all, what are they? " the answer is
this : " They are quantities for which we have defined the f undar
mental operations of numbers and, since they have the properties
of numbers, must be called numbers, just as a flower that has all
the characteristic properties of a known species is thereby deter-
mined to belong to that species." Furthermore, our operations
have been so defined that if the imaginary parts of the complex
numbers vanish and the numbers become real, the expression
defining any operation on complex numbers reduces to one defin-
ing the same operation on the real part of the number. Thus in
(1) above, if & = c? = 0, the expression reduces to
a _a
d c
COMPLEX NUMBERS
159
EXERCISES
Carry out the indicated operations.
1. (2 + V^2)(4 + V=l).
Solution : 2 + -vA^ = 2 + V2(-l) = 2 + i -v^
i + V^^ = 4 + V5(- 1) = 4 + i yg
2. 5-^-\^-i\/3.
Solution :
5 5(^^ + jV3)
8--v/lO + i4V2 + i2V6
5 V2 + i 5 V3
■V2-iVS (V2-iV3)(V2 + iV3) 2 + 3
3. (1 + i)4. 4. (1 + i)3.
Hist. Deyelcp by the binomial theorem.
= V2 + i Vs.
5. (a + iby.
7. (a; + i2/)2.
9. vr+1 • vn^.
11. (VTTI + Vr^'.
13. (Va + iVb)(Va-iVb).
15.
18.
l + iV3
l-iVs"
4
16.
1 + J
6. (VS + V^^)''.
8. (a; + J2/)2 + (a; - iv)^
10. (V3 + iV2)(V2 + iV3).
12. (aVb + icVd)(aVb-icVd).
14. (2V7+i3V8) (3^7-11072).
3
17.
l + V^^
21. (^y.
24.
19.(1^
(1 + i)«
- C-^T
20
27.
30.
V2-iV3
V3 + j\^
V3-iV2'
29
25.
21
4 + 7V^r5
31.
4 + 3iV6
64
1 + 3V^'
23.
26.
[+ i VI - a2
a — i Vl — a^
V-a+ V^
29 " + '^ c + M
32.
a — lb
1
c — id
(l+i)2^(l-i)
33.
V1+ a + i Vl - g _ Vl - a + i VI + a
Vl + o — i Vl — a Vl — a — i Vl + a
160 ADVANCED ALGEBRA
34. Find three roots of the equation x* — 1 = and represent the roots
as points on the plane.
35. Find four roots of the equation a^ — 1 = and represent the roots as
points on the plane.
36. Find six roots of a^ — 1 = and represent the roots as points on the
plane. Show graphically that the sum of the six roots is zero.
37. Find three roots of a;^ — 8 = and represent the roots as points on
the plane. Show graphically that the sum of the three roots is zero.
154. Polar representation. The graphical representation of
complex mimbers given in § 147 gives a simple graphical inter-
pretation of the operations of addition and subtraction, but the
graphical meaning of the operations of multiplication and divi-
sion may be given more clearly in another manner. We have
seen that we may represent x + iy by the point P (x, y) on the
plane. Represent the angle between OP and the X axis by 6.
This angle is called the argument of the complex number x + iy.
Represent the line OP by p. This is called
the modulus of as -|- iy. Then from the figure
T.
p
yf a:-\-iy
/\.
Ae ! ,
X i
05 = p cos 6,
(1)
y = p sin e,
(2)
f + y' = p\
(3)
Hence the complex number x -\- iy may be written in the form
x-\-iy = p (cos 6 + i sin ff), (4)
when the relations between x, y and p, 6 are given by (1), (2), and
(3). A number expressed in this way is in polar form, and may
be designated by {p, 6). We observe that a complex munber
lies on a circle whose center is the origin and whose radius
is the modulus of the number. The argument is the angle
between the axis of real numbers and the line representing the
modulus.
155. Multiplication in polar form. If we'have two numbers
p (cos + i sin ff) and jo'(cos ff' + i sin ff'), we may multiply them
and obtain
COMPLEX NUMBEES
161
By the addition tlieoiem
in Trigonometry
p (eos S + i sin 6) p'(cos 6' + i sin 6')
= pp' [(cos cos 0' — sin 6 sin 0')
+ i (sin eos S' + cos sin fl')]
= pp' [cos (fl + e') + t sin ((9 + 0')-] (1)
= ie(cos® + isin®). (2)
In this product pp' is the new modulus and 5 + 6' the new
argument. We may now make the following statement: The
product of the two numbers p (cos -\-i $in 0) and p'(cos 0' + i sin 0')
has as its modulus pp' and as its argument + 0'. Thus the
product of two numbers is represented on a circle whose radius
is the product of the radii of the circles on which the factors are
represented. The argument of the product is the sum of the
arguments of the factors.
156. Powers of numbers in polar form. When the two factors
of the preceding section (p, 0) and (p', 0') are equal, that is,
when p = p' and = 0', the expression (1) assumes the form
[p (cos + isin0)Y = p^ (cos 2 6 + i sin 2 5). (1)
This suggests as a form for the nth. power of a complex number
[p (cos + ism e)]" = p" (cos-w^ + i sin n0). (2)
The student should establish this expression by the method of
complete induction. The theorem expressed by (2) is known as
DeMoivre's theorem. Stated verbally it is as follows : The modulus
of the wth power of a number is the rath power of its modulus. The
argument of the nth power of a number is n times its argument.
EXERCISES
Plot, find the arguments and moduli of the following numbers and of
their products.
1. 1 + j V3, V3 + i.
Solation :
Let y/3 + i=p (cos » + i sin 9),
l + iVS = p' (cos e' + i sin 9^.
Then by (1), (2), (3), § 154, p = 2 ; p' = 2.
1 = 2 sin e, hence e = 30° ;
1 = 2 cos e', hence «' = 60°.
e = 30°, fi' = 60°
162 ADVANCED ALGEBEA
Thus if the product has the form B (cos + i sin 0), we have by § 155,
iJ = pp' = 4, = fl + fl' = 90°.
2. 1 + i, 2 + i.
3. (l-j)3.
4. 3 + 3 i, 2 - i Vil.
5. 2i, l-iV3.
H--'f)"
7. -l + j,-2-2i
S.H-^-^.
. 1 , iV3 -v^ , iVI
2 2 2 2
10. [2(cosl5°+isinl5°)]3. , 11. [J (cos 30° + i sin 30°)]*.
12. [f (cos 120° + i sin 120°)]2. 13. [2 (cos 135° + i sin 135°)]*.
14. [|(cosl80° + isinl80°)]3. 15. [f (cos815° + isin315°)]2.
157. Division in polar form. If we have, as before, two com-
plex numbers in polar form (p, 8) and (p', ff), we may obtain their
quotient as follows.
p (cos d + i sin &)
p'(oosd' + ism9')
pp' (cos 6 + i sin ff) (cos 6' — i sin 6')
Rationalizing, = p-(cos 0' + isin fl') (COS 6' - isin e')
§ 152 and § 153, - -^^^_^^^-_^__
Since sin^ d + cosa fl = 1, = -^ [cos (0 — 0') + i sin (9 — &y\
r
= 2? (cos ® + i sin 0).
We may now make the following statement : The quotient of
two complex numbers has as its modulus the quotient of the moduli
of the factors, and as its argument the difference of the arguments
of the factors.
158. Roots of complex numbers. We have seen that the square
of a number has as its modulus the square of the original modulus,
while the argument is twice the original argument.
This would suggest that the square root of any number, as (p, ff),
a
would have Vp as its modulus and ^ as its argument. Since
every real number has two square roots, we should expect the
same fact to hold here. Consider the two numbers
COMPLEX NUMBERS
163
Vp/cosk + *sin-j and Vp cos ( ^ + 180° j+ isinl^ + 180°
where Vp is the principal square root of p(§ 72). The square of
the first is (p, 6), by § 155. That the square of the second is the
same is evident if we keep in mind the fact that
cos {6 + 360°) = cos e
and sin {d + 360°) = sin 6.
Thus Vp (cos 6 + i sin &)
Vp I cos - + t sin ^ ) or
I + 180°) + t sin /I + 180'
Vp cos /-
)]•
The graphs of these two numbers are situated at poiuts sym-
metrical to each other with respect to the origin.
We may obtain as the corresponding expression for the higher
roots of complex numbers the following :
'6l+/<;360°\ . . /e + A360°'
cos I I + I sm
-^ p{<ios + is\Ti0)= Vp
where for a given value of n, k takes on the values 0, 1, ■••,% — 1,
and where Vp indicates the real positive nth. root of p.
EXERCISES
Perform the indicated operations and plot :
1. 2-2v^i--l+i.
Solution :
Let \-\- i = p (cos 9 + i sin S),
2 - 2 V3 i = (j'(cos e'+i sin e').
Then p = VI^+l^ = V2,
By (1) and (2), § IM,
sin e = oc® e = — =, hence e = 45°.
V2
Y
iv
^{i+i)
45°
\
X
7
\
V
P
\-
i^-i2-\f3)
164
Similarly,
Thus
ADVANCED ALGEBKA
. ^ 2-v/3 V3
sin ff' = = ,
4 2
cos «' = f = I, hence r = 300°.
2-2V3i „, _ . ._^, 4 (cos 300° + i sin 300°)
1+i
:iJ{cos® + isin®) =
V^ (cos 45° + isin 45°)
Hence by § 157, iS = -5= = 2V2, © = 300° - 45° = 255'.
V2
i,-2+i2V3
2. V-2 + 2V3i.
Let — 2 + 2v'3i = p (cos » + i sin 9).
Then (§154) /> = 4, cos 9= - f = - J,
and
e = 120°.
V-2 + 2V3i = -^4 (cos 120° + i sin 120°)
„..,.„ ■ /iF /120° + fc360°\
By § 158, = Vi l^cos {^ 31 j
l+i-\^
. . /120° + fc360'
^]
{where fe = or 1)
= 2(cos60°+isin60°) = l + iV3, when i;=0.
= 2 (cos240°+i sin240°)= — 1— iV3, when&=l.
3. VV2 + i V2.
5. -1 + i-^l-i.
2 2
4. V-l-J.
6 !_i^-
■ 2 2 ■
1 + i.
8.
1 iV3 . 1 . t
2 2 ■ 4 4'
10. -2V2-2V2i-r--2 + 2V3i.
9. 2-iVi2-^3 + 3i.
11. ■v'l (cos 15°+ isin 15°).
Solution :
8, 8_r /15°+fc-360°\ . . /15°+ fc ■ 360°\"|
VI (cos 15°+ isin 15°) = VI cos( ^ j + i sm I — ^ j
(where fc = 0, 1, or 2).
' 1 (cos 5° + i sin 5°), when fc = 0,
= • 1 (cos 125° + i sin 125°), when fc = 1,
. 1 (cos 245° + i sin 245°), when fc = 2.
12. Vi.
13. 4^i6i.
14. ■v'2 + 2V3i.
15. v'cos 330°+ isin 330°.
COMPLEX NUMBEKS
165
16. V27(oos75°+isin75°). 17. V16 (cos 200° + i sin 200°).
18. Solve the following equations and plot their roots.
(a) a;5 - 1 = 0.
Solution : a;^ = 1, or a; = V 1.
Let l = l+0-i = p(cos9 + iBin«). Then p = 1, « = 0°.
5,— -j- — . . ^„^ s/_r /0°+i;-360°\ , . . /0°+ fc.360°\
» = Vl(cosO°+ ism0°) = VI oosi 1 + isinl '
(where k takes on the values 0, 1, 2, 3, 4)
'cos 0° + i sin 0° = 1, when i: = 0,
cos 72° + i sin 72°, when k = 1,
= ■{ cos 144° + i sin 144°, when k = 2,
cos 216° + i sin 216°, when J: = 3,
. cos 288° + « sin 288°, when k = i.
These numbers we observe lie on a circle of
unit radius at the vertices of a regular pentagon.
(b) a;* - 1 = 0.
(e) a;6 - 1 = 0.
(0) a;S-l = 0.
(f ) a;« - 1 = 0.
(d) a;' - 32 = 0.
(g) a;8 - 27 = 0.
CHAPTEE XVII
THEORY OF EQUATIONS
159. Equation of the nth degree. Any equation in one variable
in which the coefRcients are rational numbers can be put in the
form
f(x) = a^x" + aia;»- > + ■•■ + a„ = 0, (1)
■where a„ is positive and «„, • • -, a„ are all integers.
The symbol/(a;) is read "/ of x " and is merely an abbreviation for the right-
hand member of the equation. Often we wish to replace x in the equation by
some constant, as a, — 2, or 0. We may symbolize the result of this substitution
l'y/(a),/(- 2), or/(0).
Thus /(6) = ao&''-|-(ii5"-l-|---- + o„.
We symbolize other expressions similarly by <t>(x), Q (x), etc.
When we speak of an equation we assume that it is in the form
of (1). This equation is also written in the form
a;» + 5ia;«-i + ...S„ = 0, (2)
where Oi = — > 0^ = —i ■■■o„ = -^-
CTq ftp (Iq
The 6's are integers only when a^, a^, ■■■, a„ are multiples of Oo-
160. Remainder theorem. We now prove the following impor-
tant fact.
Theorem. When f(x) is divided hy x — c, the remainder is
f(x) with c substituted in place of the variable.
Divide the equation (1) by x — c. Let R be the remainder,
which must (§ 26) be of lower degree in x than the divisor ;
that is, in this case, since a; — c is the divisor, R must be a con-
stant and not involve x at all. Let the quotient, which is of
degree re — 1 in a;, be represented by Q (x).
160
THEORY OF EQUATIONS 167
Then £^ = Qix)+ -^—
X — C ^' X —
Clearing of fractions,
f(x)= Q(x)(x-c)+R.
But since this equation is an identity it is always satisfied .
whatever numerical value x may have (§ 53).
Let x^ c.
Then /(c) = a^c'' + aic—^ + ••• + «„= Q (c) (c - c) + R-
But since c — c = 0, Q(c) (c — c) = 0, and
R = a^c" + diC"- 1 + . ■ . + a„ =/(c).
CoKOLLAEY. If c is a root of fix) = 0, then x — c is a factor
of the left-hand member.
For if c is a root of the left-hand member, it satisfies that
member and reduces it to zero when substituted for x. Thus by
the previous theorem we have, since
flSoC" + aiiC"-' H a„ = iJ = 0,
f(x)=Q(x)(x-c).
161. Synthetic division. In order to plot by the method of
§ 103 the equation
y = a^x" + a^x"- ^ -\ \- a„,
when the a's are replaced by integers, we should be obliged
laboriously to substitute for x successive integers and find corre-
sponding values of y, which for large values of n involves con-
siderable computation. We can make use of the preceding theorem
to lighten this labor. The object is to find, with the least possible
computation, the remainder when the polynomial fix) is divided
by a factor of form x — c, which by the preceding theorem is the
value of f(x) when x is replaced by c, that is, the value of y
corresponding to a; = c. For illustration, let
f(x) =2x*-3x'-\-x^ — x-9 a,ndL0 = 2.
168 ADVAIfCED ALGEBRA
By long division we have
a!-2| 2a!^-3a:°+ a;^ - x- 9 |2a:° + a;" + 3a; + 5
2a!*-4a;=
la;»
+
x"
la:»
—
2x^
Zx^-
X
3x^-
6x
5x-
- 9
5x-
-10
+ 1
We can abbreviate this process by observing the following
facts. Since x is here only the carrier of the coefficient, we may
omit writing it. Also we need not rewrite the first number of
the partial product, as it is only a repetition of the number
directly above it in full-faced type. Our process now assumes
the form
1-2| 2 -3 + 1-1- 9 |'2 + 1+3 + 5
-4
+ 1
-2
+ 3
+ 5
-10
+ 1
Since the minus sign of the 2 changes every sign in the partial
product, if we replace — 2 by + 2 we may add the partial prod-
uct to the number in the dividend instead of subtracting. This
is also desirable since the number which we are substituting for
X is 2, not — 2. Thus, bringing all our figures on one line and
placing the number substituted for x at the right hand, we have
2-3 + 1-1- 9[2
+ 4 + 2 + 6 + 10
2+1+3+5+ 1
THEORY OF EQUATIONS 169
We observe that the figures in the lower Une, 2, 1, 3, 5,, up to
the remainder are the coefiicients of the quotient 2x^ + x^ + Sx-\-5.
EULE FOE SYNTHETIC DIVISION. Write the coefficients of the
folynomial in order, supplying when a coefficient is lacking.
Multiply the number to he substituted for x by the first coeffi-
cient) and add (algebraically) the product to the next coefficient.
Multiply this sum by the number to he substituted for x, add to
the next coefficient, and proceed until all the coefficients are used.
The last sum obtained is the remainder and also the value of
the polynomial when the number. is substituted for the variable.
162. Proof of the rule for synthetic division. This rule we now
prove in general by complete induction. Let the polynomial be
a^x" + OiX""' + a^x"'^ + ■■• + «„•
Let the number to be substituted for x be a.
First. Let n = 2. Carry out the rule on a^x^ + a^x + eia.
We have , , ,
aa +«! H-ttz [a
+aoa -\-{a^ + ai)a
«o; + So* + «i) + (»o* 4- «i) « + aa = afflc^ + %« + a^.
Second. Assume the validity of the rule for n = m, and prove
that its validity for w = to + 1 follows. Assume then that the rule
carried out on
fix) = flSo*'" + aia;"-^ -\ h «„
affords the remainder
a„a'^ + fflia"-! + •■• + »« =/(«)•
Now the polynomial of order to + 1 is
ajaj^+i + flSia;'" H \- a^x -\- a„+i = x -/(a;) + a„+i.
Hence the next to the last remainder obtained by applying the
rule to this polynomial would be f(a), since the succession of
coefficients is the same for both polynomials up to a„^.i. By
the rule the final remainder is obtained by multiplying the expres-
sion just obtaiued, in this case f{a), by a and adding the last
coefficient, in this case a„+i. This affords the final remainder
«•/(«) + a„+i = a(,a'» + i + Oja™ + ... + «„« + as^+j.
170
ADVANCED ALGEBRA
(c) 4x»-7a;-87 by p: - 3.
(e) x3 + 4a;2 - 7x - 30 by a; + 3.
EXERCISES
1. Prove by complete induction that the partial remainders up to the final
remainder obtained in the process of synthetic division are the coefficients of
the quotient oif(x) Toy x — a.
2. Perform by synthetic division the following divisions.
{a) x»-Tijfi-6x + 12 hy x-i. '
Solution: 1-7- 6 + 72 [4
4-12-72
1 _ 8 - 18
Quotient = x^-Sx-lS.
(b) a;»-9a; + 10 by k - 2.
(d) s' + 8x2 - 4s - 32 by a; - 2.
(f) xs - 6x2 + 11 X - 6 by X - 1.
(g) X* - 16x» + 86x2 - 176x + 105 by x2 _ gx + 7.
Hint. Sincex2— 8a!4-7=(x— 7)(x— 1), divide by x— 7 and the quotient by x—1.
(h) x« + 1 by X + 1. (i) x9 - 1 by X - 1.
(i) X* + x' - X - 1 by x2 - 1. (k) x6 - 2x8 - 4x by X - 3.
(1) x6 - 2x' - 4x - 1 by X + 2. (m) 4x' - 6x2 - 2x - 1 by x - 3.
(n) 2x* + 6x8 _ 37a;a +44x + 84 by x2 + 5x - 6.
163. Plotting of equations. We can novsr form tlie table of
values necessary to plot an equation of the type
a„a;» + a^x"- ^ -\ \- a„_iX + a„ = y.
Example. Plot x' + 4 x2 — 4 = 2/.
l + 4 + 0-4[l
+1+5+5
1+5+5+1
1 + 4 + 0-41-1
-1-3+3
1+3-3-1
l + 4 + 0-4|-2
-2-4+8
1+2-4+4
1 + 4 + 0-41-3
-3-3+9
+1-3+5
1 + 4 + 0-41-4
-4+0+0
1+0+0-4
THEOEY OF EQUATIO^TS 171
In this figure two squares are taken to represent one unit of x. A single
square represents a unit of y.
By an inspection of the figure it appears that the curve crosses the X axis
at about x = .8, x = — 1.2, and a; = — 3.7. Thus the equation fory = has
approximately these values for roots (§ 110).
164. Extent of the table of. values. Since the object of plot-
ting a curve is to obtain information regarding the roots of its
equation, stretches of the curve beyond all crossings of the X axis
are of no interest for the present purpose. Hence it is desirable
to know when a table of values has been formed extensive enough
to afEord a plot which includes all the real roots. If for all values
of X greater than a certain number the curve Ues wholly above
the axis, there are no real roots greater than that value of x.
By inspection of the preceding example it appears that if for
a given value of x the signs of the partial remainders are all
positive, thus affording a positive value of y, any greater value
of X will afford only positive partial remainders and hence only
positive values of y.
Thus when all the partial rem.ainders are positive no greater
positive value of x need be substituted.
Similarly, when the partial remainders alternate in sign begin-
ning with the coefficient of the highest power of x, no value of x,
greater negatively, need be substituted.
In plotting, if the table of values consists of values that are
large or are so distributed that the plot would not be well propor-
tioned if one space on the paper were taken for each unit, a scale
should be so chosen that the plot will be of good proportion,
that is, so that all the portions of the curve between the extreme
roots shall appear on the paper, and the curvatures shall not be
too abrupt to form a graceful curve. This was done, for example,
in the figure, § 163.
EXERCISES
Plot and measure the values of the real roots of the equations when y — 0.
1. x'-tx-Q^y. 2. x«~7x+ 5 = y.
3. 1x'-9x-6 = y. 4. a;8 - 31a; -I- 19 = 2/.
5. x^-12x-U = y. 6. ix'-lSx + 6 = y.
172 ADVANCED ALGEBRA
7. a;8 _ 12a; - 16 = y. 8. x^ - 45a; + 152 = y.
9. Ki - 2a;s - a; + 2 = 2/. 10. 8x' - 18a;2 + 17a; - 6 = y.
11. x* - 17a;2 + a; + 20 = 2/. 12. a* - 4x3 + 9a;2 _ 8x + 14 = y.
13. 18x3- 36x2 + 9x + 8 = y. 14. x* + 5xs + 12x2 + 52x-40 = 2/.
15. x*-2x3-7x2+19x-10=2/. 16. x* - 6x8 + Sx^ + 26x -24 = j/.
17. 6x* - 13x3 + 20x2 _ 37a; + 24 = y.
165. Roots of an equation. In the case of the linear and
quadratic equations we have been able to find an explicit value
of the roots in terms pf the coefficients. Such processes are prac-
tically impossible in the case of most equations of higher degree.
In fact the proof that any equation possesses a root lies beyond
the scope of this book, and we make the
Assumption. Every equation possesses at least one root.
This is equivalent to the assumption that there is a number,
rational, irrational, or complex, which satisfies any equation.
166. Number of roots. We determine the exact number of
roots by the following
Theorem. Every equation of degree n lias n roots.
Given the equation /(a;) = aoCc" + aio;""^ + ••■ + «„ = 0.
L'et «] (see assumption) be a root of this equation. Then (p. 166)
a; — «! is a factor of the left-hand member, and the quotient of
f(x) by a; — «! is a polynomial of degree ?i — 1. Suppose that
aoa;» + aj.x''-' + ■■ ■ + a„ = ao(x- aj) (a;"-i -t- hx"-^ + ■■■ + 6„_i).
By our assumption the quotient a;"~^ -|- Jia;""^ ^ . . . ^ 5^_ ^ =
has at least one root, say aj, to which corresponds the factor
X — a^. Thus
/(a;) = a„(a; - a^ (x - a,) (a;"-" -f- c^x"-' + ■■■ + c„_,).
Proceeding in this way we find successive roots and corre-
spoiiding linear factors until the polynomial is expressed as the
product of n linear factors as follows :
f(x) =ao(x — ai) (a; - aa) • ■ • (a; - a„) = 0,
where the roots are oti, otj, • ■ •, ar„.
Remark. This theorem gives no information regarding how many of the roots
may he real or imaginary. This depends on the particular values of the coefficients.
THEORY OF EQUATIONS 173
CoEOLLAEY. Any polynomial in x of degree n may be expressed
as the product of n linear factors of the form x — a, where a is
a real or a complex number.
It should be noted that the roots are not necessarily distinct.
Several of the roots and hence several of the factors may be
identical
If f(x) is divisible by (x — a{f, that is, if aj = a^, we say that
oTi is a double root of the equation. Similarly, if f{x) is divis-
ible by (x — ari)'", a-^ is called a multiple root of order / . When
we say an equation has n roots we include each multiple root
counted a number' of times equal to its order.
Theoeem. An equation of degree n has no more than n
distinct roots.
Let/(a;) = aox" -| f- a„ = have the roots oTj, a^,- ■ ■, a„. Write
the equation in the form
ao(x — ai) ■ • • (a; — «„) = 0.
If r is a root distiact from a^, •••,«„, it must satisfy the equation
and
ao(r — a{)---{r — a„) = 0.
Since this numerical expression vanishes one of its factors
must vanish (§ 6). But r ^ a^, thus r — a^ ^ 0. Similarly, no
one of the binomial factors vanishes. Thus (§ 5) a^ = 0, which
contradicts the hypothesis that the equation is of degree n.
This theorem may also be stated as follows :
COEOLLAEY I. If an equation «„»" + a.^x" ~' H \- a^= of
degree n is satisfied by more than n values of x, all its coefficients
vanish.
The proof of the theorem shows that if the equation has n + 1
roots, oso = 0- W^ should then have remaining an equation of
degree n—1, also satisfied by ra +1 values of x. Thus the coeffi-
cient of its highest power in x vanishes. Similarly, each of the
coefficients vanishes.
174 ADVANCED ALGEBRA
Corollary II. If two 'polynomials in one variable are equal
to each otJier for every value of the variable, the coefficients of
like powers of the variable are equal and conversely.
Let aox" + aiK"" ^ -\ h «„ = h^" + 61 as""^ + i- b„
for every value of x.
Transpose, (uo — bo)x'' + ■ ■ ■ + a„ — b„ = 0.
By Corollary I, , ag — bo = 0, or tto = bo,
ai — 5i = 0, or a^ = bi,
a„-b„ = 0, or a„ = b„.
167. Graphical interpretation. -The graphical interpretation
of the theorems of the preceding section is that the graph of an
equation of degree n cannot cross the X axis more than n times.
Since each crossing of the X axis corresponds to a real root, there
will be less than n crossings if the equation has imaginary roots.
168. Imaginary roots. We now show that imaginary roots
occur in pairs. This we prove in the following
Theorem. If a + ih is a root of an equation with real coeffi-
cients, a — ib is also a root of the equation.
If a. + ib is a root of the equation cso*" + aix^~ ' + ••• + a„ = 0,
then x —(a + ib) is a factor (p. 166). We wish to prove that
X — (a — ih) is also a factor, or what amounts to the same thing,
that their product
[x — (a + i6)] [a; — (a — iFf\ = [(a; — a)— t6] [(a; — a) + ib"]
= (x - ay + b^
is a factor of f{x). Divide f{x) by (x — a)' + b^ and we get
f{x) = Q{x) [(x - ay + 6=] + « + r', (1)
where r and r' are real numbers. This remainder rx + r' can be
of no higher degree in x than the first, since the divisor
{x - a)^ + b"
THEORY OF EQUATIONS
175
is only of the second degree (§ 26). Now this equation (1) being
an identity is true whatever value is substituted for x, as, for
instance, the 'root of f{x), a -\- ib. Substituting this value for x,
we get
f(a + ib)=0 = Q(a + ib) [(a + ih - af + b^'\ + r (-a + ib) + r',
or (p. 33) and (p. 3) = + ra + r' + irb,
or (p. 156) ra + r' = 0, (2)
rb = 0. (3)
Since S ?fc 0, by (3), p. 3, r = 0.
Also from (2), r' = 0.
Hence rx + r' = 0.
Consequently there is no remainder to the division of f(x) by
(x — ay + b^, and hence if a + t J is a root oif(x), a — ib is also a
root.
CoEOLLAEY. Every equation of odd degree with real coeffi-
cients has at least one real root.
The roots cannot all be imaginary, else the degree of the equa-
tion would be even by the preceding theorem.
169. Graphical interpretation of imaginary roots. When we
plot the equations
y
—
a;»
+
4
a;2
4
(
1)
f
\"
/
N
/
V
/
\
/
j
)
f
/
V
J
/
\
I
i
/
/
/
/
^
/
y = x^ + 1x^-1 (2),
176
ADVANCED ALGEBRA
2/ = a:' + 4a;= (3),
y
^
x"
+ 4
1
x"
+1 (4),
iirA 1
^
<"
f
/
N
/
/
/
j
\
1
\
/
\
/
V
/
\
f
V
J
1
4
—
—
—
—
1
we see that corresponding to the increase of the constant term is
a corresponding elevation of the curve with respect to the X axis.
In fact in each case the curve is the same, but the value of y is
gradually increased. In (1) and (2) we have three real roots, in
(3) the curve touches the X axis, and in (4) we have only one
real root. As the elbow of the curve is raised and fails to intersect
the X axis a pair of roots cease to be real, and since a cubic equa-
tion always has three roots, a pair of roots become imaginary.
Thus we have the
Peinciple. Corresponding to every elbow of the curve that
does not intersect the X axis there is a pair of imaginary roots
of the equation.
The converse is not always true. It is not always possible to
find as many elbows of the curve which do not meet the X axis
as there are pairs of imaginary roots.
EXERCISES
Plot the following equations and determine from the plot how many roots
are real.
1. X* - 1 = y. 2. x6 - 2 = y. Z. 1^ -x-l=y.
4. a!* + l = y. 5. K* -fa; + 1 = 2/. 6. a;* + 2x2 -|- 2 = y.
7. x» - 3x2 - X + 1 = 3/. 8. x' - 2x2 + 4x - 1 = y.
9. 2x8 -I- 3x2 + 5x -H 6 = y. lo. x' - 3x2 - 4x - 5 = y.
THEORY or EQUATIONS 177
170. Relation between roots and coefficients. If we write the
expression (Corollary, p. 173)
a;» + Sia;" - ' + . . . + }„ = (a; - ^i) (x - jSa) • ■ ■ (a; - ^„) ,
and multiply tlie factors, we obtain by equating coefficients of like
powers of x (p. 174) relations between tlie roots and tbe coeffi-
cients. Take for example ?i = 3.
x' + b^x^ + b^x + b, = (x- PO (x - 13,) (X - jSa)
= »» - (|8i + ^, + fi,-) x^ + (J3,^, + 13,13, + 13^13,) x - I3,fi,ps = 0.
Hence 6i = - (/8i + A + /8s),
bz=-P^PiPz.
This suggests the
Theorem. Tlie coefficient of »""' is equal to the sum of the
roots with their signs changed.
The constant term, is equal to the product of the roots with
their signs changed.
In general the coefficient of of" is equal to the sum of all
possible products of r of the roots with their signs changed.
We prove this theorem by complete induction. •
First. We have already established the theorem for equations
of degree two on p. 106 and for equations of degree three above.
Second. Assume the theorem for ?i = m. That is, if
a;™ + SicC"-^ + • ■ • + 6„ = (a: - /3i) (» - A) ■ • • (a: - /8„), (1)
we assume that b„ the coefi3.cient of x^'', is the sum of all possi-
ble products of r of the numbers — Pi, — h ■ ■ ■, — /3„.
Multiply both sides of (1) by x — ;8„+i. Denote the result by
^»+i + j,^^,™ + . . . + jr^^^ = (^ _ ^^)(^ _ ^^) . . . (a; _ ^„^,). (2)
The term in a;"+'-'" in this equation is obtained by multiplying
the terms S^a;'""'" and b^_^x'"+'^"' in (1) by x and — /3„^.i respec-
tively. That is, in (2)
JV = Jr + &r-i(-)8™+0- (3)
Now all possible products of r of the quantities — ;8i, —Pa
■ ■; — /3„+i may be formed as follows : (1) Neglect — /8„+i, and
form all possible products of r of those remaining. The sum of
178 ADVANCED ALGEBRA
these is b^ (2) Form all possible products of r — 1 of — )8i
— Aj • • •> — Pm> not including — /3„ + i, and multiply eacli product
^y -~ Pm+\- Add all the products obtained. This process, it is
observed, is precisely that indicated by (3).
Remark. It is noticed that in the rule the signs of the roots are always changed
before forming any term. This does not involve any change when r is an even
number, but is included in the rule for the sake of uniformity.
Corollary. Every root of an equation is a factor of its con-
stant term.
171. The general term in the binomial expansion. On p. 129
we gave an expression for the {r + l)st term of the binomial ex-
pansion, the validity of which we now establisL In (1), § 170,
let ;8i = /Sa = ■ ■ ■ = p„. Denote this common value by — a. The
expression (1) becomes, on writing n in place of m,
a;" + 5ia;"-" H \-h^=(x + ay.
By the theorem in § 170, S, is the sum of all possible products
of r of the negative roots. Since there are
n(ji — V)--- {n — r + 1)
"•'•"" r\
such products, .and since the roots are now identical, we obtain
n(n-V).--(n-r + l) ^„_^^^
r\
as the form of the (r + l)st term of the expansion of (x + a)".
172. Solution by trial. Since by the previous corollary every
root of an equation is a factor of its constant term, we may in
many cases test by synthetic division whether or not a given equa-
tion has integral roots. Thus the integral roots of the equation
a;«_8a!»-f4a!2-f 24a; -21 = q^
must be factors of 21.
We try -}- 1 by synthetic division,
l_8 + 4-f24- 2111
-H-7- 3-H21
l-7-3-h21
Thus 1 is a root of (1), and the quotient of the equation by
'^-'^^ a;»-7a;»-3a! + 21 = 0. (2)
THEOBY or EQUATIONS 179
If this equation has any integral root it must be a factor of 2L
We try + 3 by synthetic division,
1-7- 3 + 21[3
+ 3 _ 12 - 45
1 _ 4 _ 15 _ 24
Thus 3 is not a root. We try + 7,
1 _ 7 _ 3 + 21[7
+ 7 + 0-21
1+0-3
Thus 7 is a root, and the .remaining roots of ,(1) are the roots of
a;» - 3 = 0,
that is, X =± Vs.
Hence the roots of (1) are + 1, + 7, ± Vs.
EXERCISES
Solve by trial :
1. a;3 - 7 a;2 + 50 = 0. 2. a;3 - 9a; + 28 = 0.
3. a;8 - 36a; - 91 = 0. 4. cc^ + 9a; + 26 = 0.
5. a;8 - 19a; + 30 = 0. 6. a;3 - 27a; - 54 = 0.
7. a;* + 2a:2 - 23a; + 6 = 0. 8. a;3 - 6a;2 + llx - 6 = 0.
9. a;s - 2a;2 -11a; + 12 = 0. 10. x^ - 8x^ + 19x - 20 = 0.
11. a;8 + 9x2 + 27a! + 26 = 0. 12. x* - Sx^ + 8x2 + 40x - 32 = 0.
13. x* - 13x2 + 48x- 60 = 0. 14. a;* - 8x'- 34x2 + 18x+168 = 0.
15. X* + 8x' - 7x2 - 50x + 48 = 0.
16. x* - 3x8 - 5a;2 + 29x - 30 = 0.
17. X* - 6x8 + 13x2 - 30x + 40 = 0.
18. x4-8x8 + 21x2-34x + 20 = 0.
19. x* - 12x' + 43x2 _ 42x + 10 = 0.
173. Properties of binomial surds. A binomial surd is a;
number of the form a ± Vs, where a and b are rational numbers,
and where b is positive but not a perfect square.
Though we have not explicitly defined what we mean by the sum of an
Irrational number and a rational number, we shall assume that we can
operate with the binomial surd just as we would be able to operate if 6 were
a perfect square.
180 ADVANCED ALGEBRA
Theoeem 1. If a binomial surd a + V& = 0, then a=0 and
b^O.
If a + Vs = and either a = or 5 = 0, clearly both must equal
zero. Suppose, however, that neither a nor b equals zero. Then
transposing we have a = — Vi, and a rational number would be
equal to an irrational number, which cannot be. Hence the only
alternative is that both a and b equal zero.
Theorem II. If two binomial surds, asa + V& and c + V5, are
equal, then a = c and b = d.
Let , a + ■Vb = o ri- 'Vd.
Transposing e, a — c + Vo = V^. (1)
Square and we obtain
(a - cy+ b + 2(a-c)Vb = d,
or
(a - cy + b ~ d + 2(a - e) Vb = 0.
Thus, by Theorem I, either b = 0, which is contrary to the defi-
nition of a binomial surd, or a — o = 0, that is, a = c. In the
latter case (1) reduces to Vs = Vd, oib = d, and we have a = o and
b = d, which was to be proved.
a + Vs and a — Vj are called conjugate binomial surds.
Theorem III. If a given binomial surd a + V& is the root of
an equation with rational coefficients, then its conjugate is also a
root of the same equation.
The proof of this theorem, which should be performed in writ-
ing by each student, may be made, analogously to the proof of
the theorem on p. 174.
174. Formation of equations. If we know all the roots of an
equation, we may form the equation in either one of two ways
(see p. 167 and p^ 177).
First method. If a^, a^---, a„ are the given roots, multiply
together the factors x — a^,---,x — a^.
Second method. From the given ^ roots form the coefficients
hy the rule on p. 177.
THEORY OF EQUATIONS 181
If the equation and all but one of its roots are known, that
root can be found by the solution of a linear equation obtained
from the coefilcient of the second or the last term. If all but two
of its roots are known, the unknown roots may be found by the
solution of a pair of simultaneous equations formed from the
same coefficients.
In the solution of the following exercises use is made of the
theorem on p. 174, Theorem III, p. 180, and the various relations
between the roots and the coefficients.
EXERCISES
1. Form the equations which have the following roots. Check the process
by using both methods of § 174
(a) 2, - 3, 1.
Solution : '
First method. (x - 2){x + S)(x -1) = x> -1 x + 6.
Second method. Let the equation be
x' + bix^ + hiX + 63 = 0.
Then, by § 170, 61 = - (2 - 3 + 1) = 0,
62 = -6 + 2-3 = -7,
6s = -2-3.-l = 6.
The equation then is x? — 1x + & = 0.
(b) 1, 2, 3. (c) 2, 2, 2, 2.
(d) 3, 1, 1, 0. (e) 1, 0, 0, 0.
if) ±V2, ±J. (g) 2,4, -5.
(h) 2, - 3, 1, 0. (i) 2, 3, - 5.
a)7, V5, -VS. (k) 1, 2, - I, - i.
(1) 3, 1 + i, 1 - i (m) - 4, - 3, 3 ± VS.
(n) 1 ± i, - 1 ± i (0) 2, V=^, - V=~3.
(p) -1,2,3,-4. (q)2|, 3J, -U, - 2J.
(r) ± VS, ± i V7. (s) - 5, 2 + VS, 2 - VS.
(t)±l,±^. (u) 1±|^, ^if2^.
(.) 3, Zl±^, ^1^. (w) - 1, l±i^^ lzi2^.
182
ADVANCED ALGEBRA
2. The equation x* + 23? —7 x^ — 8x + 12 = has- two roots — 3 and
+ 1. Find the remaining roots.
Solution: Let the unknown roots be a and 6.
Then, by § 170, -a-6 + 3-l = 2,
-3a6 = 12.
Solving for a and &, we obtain a = — 2 or +2,
6 = + 2 or -2.
3. x' — 7x + 6 = has the roots 2 and 1. Find the remaining root.
4. x' — 3a; + 2 = has the root 1. Find the remaining roots.
5. x' — 18 X — 35 = has the root 5. Find the remaining roots.
6. Two roots of x* — 35 x^ + 90 x — 56 = are 1 and 2. Find the remain-
ing roots.
7. The roots of x» - 6 x^ - 4 x + 24 = are in A.P. Find them.
8. The two equations xS-6x2 + llx-6 = and x^ -Ux^ + 63x - 90 =
have a root common. Plot both equations on the same axes, and find all the
roots of both equations.
9. Determine the middle term of the equation whose roots are — 2,
+ 1, 3, — 4 without determining any other term.
10. "What is the last term of the equation whose roots are — 4, 4, ± V— 3 ?
11. One root of x* — ix' + tx^ + 2x + 52 = 0is3 — 2i. Find the remain-
ing roots.
12. One root of X* - 4 x8 + 5 x2 + 8 X — 14 = is 2 + i V3. Find the others.
13. Plot the following equations, determine all the integral roots, and
find the remaining roots by solving.
(a) X* - 6 xs + 24x - 16 = 0.
FA
In this plot two squares on the X axis represent a unit of x, while one
square on the Y axis represents ten units of y. The integral factors are
X — 2 and x + 2, since ± 2 are roots, that is, are values of x for which the
THEORY OF EQUATIONS 183
curve is on the X axis. To find the quotient of our equation we first divide
synthetioally by 2, and then the quotient by — 2, using the principle given
in § 161.
1-6+ + 24-16[2
+ 2 - 8-16 + 16
1-4- 8+ 8 |-2
- 2 + 12 - 8
1-6+ 4
Thus the quotient of the polynomial and (a; - 2) (« + 2) is s" ^ 6 a; + 4.
Solving the equation
a;2_6a; + 4 = 0,
we obtain the two remaining roots, a; = 3 ± V5. These remaining roots
might also be found by the method of exercise 2.
(b) a;«-5a;-12 = 0. (o) a;'- 8a;2 + 7 = 0.
(d) a;8 - 7a;2 + 50 = 0. (e) a;' - 8a;2 + 13a;- 6 = 0.
(f) a;8 - 6a;2+ 7a; - 2 = 0. (g) a;8 + 3a;2 + 4a; - 24 = 0.
(h) a;*-3a;3 + 7a;2-21a; = 0. (i) a;*- 3a;3 - 7a;2 + 27k - 18 = 0.
(j) a;*-9a;3 + 21a;2-19a; + 6 = 0.
(k) How many imaginary roots can an equation of the 5th degree have ?
(1) a;' — aa;2 + 6a; + c = has two roots whose sum is zero. What is the
third root ? What are the two roots whose sum is zero ?
(m) a;8 + a;2 + 6a; + c = has one root the reciprocal of the other. What
are the values of the roots ?
(n) a;' — 4x2 + aa; + 52 = has the sum of two roots equal to zero. What
must be the values of a and 6 ?
(o) a;* — 3 x' + 6a; + 9 = has the sum of three of its roots equal to zero.
What must be the value of 6 ?
175. To multiply the roots by a constant. Suppose vre have
the equation ,
fix) = a„a;» 4- ai*"- ' + • • • + a„ = 0, (1)
-whose roots are a^, a^, ■■■, ar„. An equation of this type for
values of n greater than 2 is usually not solvable by elementary
methods. It often happens, however, that by changing its form
slightly we may obtain an equation one or more of whose roots
we can find. We shall see that if an equation has rational roots
we may always find them if we change the form of the equation
as indicated on the following page.
184 ADVANCED ALGEBRA
We seek to form from (1) an equation whose roots are equal
to the roots of (1) multiplied by a constant factor, as k. Thus
the equation we seek must have the roots ka^, ka^, kug. We
carry out the proof, -which is perfectly general, on the equation
of the third order
f(x) = a^x' + Oix' + a^x + aj = 0,
whose roots are a^, a^, as. The equation that we seek must have
roots kui, ka^, teg. Since now (§ 53) f{x) = is satisfied by
a, where a stands for any one of the roots, that is, since f(a) = 0,
evidently /(t) = is satisfied by ka, that is,
/(¥)=/«-»•
Hence we obtain an equation that is satisfied by kui, ka^, ka^,
if in/(a;) we let x = -•
The required equation is then
8 /Y 5/2
/(l)=^
or, multiplying by A',
ttos' + ka^z^ + k^a^z + /c'ag = 0.
This affords the general
EuLB. To multiply the roots of an equation hy a constant k,
multiply the successive coefficients beginning with the coefficient of
a^'^ hy k,lc^, • ■ -jk^ respectively.
In performing this operation the lacking powers of x should be
supplied with zero coefficients.
Example. Multiply the roots of 2 a;' — 3x + 4 = by 2.
Multiply the coefficients by the rule above,
2xa + 2-0a;2-4-3x+8-4 = 0.
Simplifying, spS - 6 a; + 16 = 0.
THEORY OF EQUATIONS 185
When an equation in form (2), p. 166, has fractional coefficients,
an equation may be formed whose roots are a properly chosen
multiple of the roots of the original equation and whose coefB.-
cients are integers.
CoEOLLAKY L When Je is a fraction this method serves to
divide the roots of an equation hy a given number.
Corollary II. When k = — 1 this method serves to form an
equation whose roots are equal to the roots of the original equa-
tion but opposite in sign. This is equivalent to the statement
that f{—oS) = has roots equal but opposite in sign to those
off{x) = 0.
EXERCISES
1. Form the equation whose roots are three times the roots of
a;4_6a;3- 1 + 1 = 0.
Solution : Supplying the missing term in the equation, we have
cc* - 6a;3 + 0a;2 - a; + 1 = 0.
Since fc = 3, we have hy the rule
a;4 _ 3 ■ 6x8 + 9 • 0x2 - 27 • a; + 81 = 0,
or x*-18x3 -27x + 81 = 0.
2. Find the equation whose roots are twice the roots of
x4 + 3x3-2a; + 4 = 0.
3. Find the equation whose roots are one half the roots of
x=-2x2 + 3x-4 = 0.
4. Find the equation whose roots are two thirds the roots of
x8-4x-.6 =0.
5. By what may the roots of the following equations be multiplied so that
in the resulting equation the coefficient of the highest power of x is unity
and the remaining coefficients are integers ? Form the equations.
(a) 3xS-6x + 2 = 0.
Solution: We wish to bring into every term such -a. factor that all the
resulting coefficients are divisible by 3.
Let k = %.
Supply the lacking term,
3x» + 0x2-6x + 2=0.
By rule, 3x' + 3 • 0x2 - 9 -Gx + 27 ■ 2 = 0.
Dividing by 3, x= - 18 x + 18 = 0.
186: ADVANCED ALGEBRA
(b) x8 + ^-|-l = 0. (c)x»-J = 0.
(d)x'> + |x» + Jx + ^ = 0. (e)x* + |-| + l = 0.
(f) 2x' - 3x2 - X + 4 = 0. (g) 3x4 - Sxi! - 4x + 1 = 0.
(h) x4-6x8-2x2+ J = 0. (i) 16x4-24x» + 8x2-2x + l = 0.
6. Form equations whose roots are the negatives of the roots of the fol-
lowing equations.
(a) x»-4x + 6 = 0.
Solution : Supply the lacking term,
,x3+ 0x2-4x + 6 = 0.
Changing signs we obtain by Corollary II
a;3_ox2-4x-6 = 0,
or x' — 4 X — 6 = 0.
(b) x'-2xii-4x = 0. (c) x*- 3x2 + 1 = 0.
(d) X* - 2x8 + x" + 2x - 1 = 0. (e) x' + 3x2 + 7x - 13 = 0.
7. What effect does changing the sign of every term of the member
involving x have on the graph of an equation?
8. What is the graphical interpretation of the transformation which
changes the signs of the roots of an equation, that is, what relation does
the graph of th^ equation before transformation' bear to the graph of the
equation after transformation (a) when the degree is an even number,
(b) when the degree is an odd number?
9. If 4x*-16x'- 85x2 + 4x + 21 = has as two roots -J and -3,
what are the roots of 4x* + 16x' — 85x2 — 4x + 21 = ?
10. If a root of x8 - 11 x2 + 36x - 36 = is 2, what are the roots of
x» + 11 x2 + 36 X + 36 = ?
176. Descartes' rule of signs. A pair of successiye like signs in
an equation is called a continuation of sign. A pair of successive
unlike signs is called a change of sign.
In the equation
2a;<-3a:S + 2a;2 + 2a;-3=0 (1)
are one continnation of sign and three changes of sign. This may be seen more
clearly by writing merely the signs, + — + + —.
Let us now inquire what effect if any is noted on the number
of changes of sign in an equation if the equation is multiplied by
THEORY OF EQUATIONS 187
a factor of the form x — a vhen a is positive, that is, when the
number of positive roots of the equation is increased by- one.
Let us multiply equation (1) by a; — 2. We have then
2a!*-3a;» + 2a;'' + 2x -3
x-2
2a;5 - 3a;* + 2£B» + 2a!2 - 3a!
-4a:* + 6a;° — 4a!'-4a! + 6
2a;=-7a;'' + 8a;'-2a;'-7a; + 6
In this expression the succession of signs is + — + — — +,
in which there are foiir changes of sign, that is, one more change
of sign than in (1). If an increase in the number of positive roots
always brings about at least an equal increase in the number of
changes of sign, there must be at least as many changes of sign in
an equation as there are positive roots. This is the fact, as we
now prove.
Descaktes' eule of signs. An equation f(x)=0 has no more
real positive roots than /(x) has changes of sign.
Illustration. In the equation of degree one a; — 2 = there
is one change of sign and one positive root. In the ease of a linear
equation there is no possibility of more than one change of sign.
In the quadratic equation a;^ + 2a; + l = there is no change of
sign, and also no positive root since for positive values of x the
expression a;^ + 2 as + 1 is always positive and hence never zero.
In the equation a;^ + 2a! — 3 = 0, we have one change of sign,
and one positive root, + 1.
We shall prove this general rule by complete induction.
First. We have just seen that the rule holds for an equation
of degree one.
Second. We assume that the rule holds for an equation of
degree m, and prove that its validity for an equation of degree
m + 1 follows. We shall show that if we multiply an equation
of degree m by a; — a, where a is positive, thus forming an equa-
tion of degree m + 1, the number of changes of sign in the new
equation always exceeds the number of changes of sign in the
188
ADVANCED ALGEBRA
original equation by at least one. That is, the number of changes
of sign increases at least as rapidly as the increase in the number
of positive roots when such a multiplication is made.
Let/(a3) = represent any particular equation of the nth degree.
The first sign of f(x) is always +. The remaining signs occur in
successive groups of + or — signs which may contaia only one
sign each. If any term is lacking, its sign is taken to be the same
as an adjacent sign. Thus the most general way in which the
signs of f{x) may occur is represented in the following table,
in which the dots represent an indefinite number of signs. The
multiplication of f(x) by a; — a is represented schematically, only
the signs being given.
An + signs
All - signs
All + signs
All - signs
Further
groups
AU-
signs
/w
+ •••• +
+•■•■+
_ _
+ •••• +
_ . .
• —
X — a
+
—
xf(x)
+ + ■■■ +
+ +■•■ +
. . _
+ +■■• +
- ^/(x)
-+•■• +
+
-+••■ +
+
- +
+ +
(x-a)f(x)
+ ±---±
-±---±
+ ±---±
-±---±
+ ± ±
-it-
••± +
The ± sign indicates that either the + or the — sign may occur
according to the value of the coefficients and of a. The verti-
cal lines denote where changes of sign occur in f(x). Assuming
that all the ambiguous signs are taken so as to afford the least
possible number of changes of sign, even then in (x — a)f(x)
there is a change of sign at each or between each pair of the
vertical lines, and in addition, one to the right of all the vertical
lines. Thus as we increase the number of positive roots by one
the number of changes of sign increases at least by one, perhaps
by more.
The only possible variation that could occur in the succession
of groups of signs in f{x), namely, when the last group is a
group of + signs, does not alter the validity of the theorenu
TPIEORY OF EQUATIONS 189
We illustrate tlie foregoing proof by the following particular
example.
Let f{x) = x^ -4.x'-x-\-2, and let a = 2.
4 changes
Multiply /(x),
1+0-4+0-1+2
by x-2,
1-2
xf{x),
1+0-4+0-1+2
- 2/(x),
-2-0+8-0+2-
-4
(X - 2)f{x),
1-2-4+8-1+4-
-4
5 changes
177. Negative roots. Since /(—a) has roots opposite in sign
to those of /(x) (p. 185), we can state
Descartes' eule of signs for negative roots. f{x) has nc
more negative roots than there are changes in sign in /(— x).
If by Descartes' rule it appears that there cannot be more than
a positive roots and b negative roots, and if a + b<n, the degree
of the equation, then there must be imaginary roots, at least
n — (a + b) in number.
EXERCISES
1. Prove Descartes' rule of signs for x^ + bx + c^O directly from tlie
expression for 6 and c in terms of the roots (see § 115).
2. Find tlie maximum number of positive and negative roots and any
possible information about imaginary roots in the following equations.
(a) a;' + 2x2+i = o.
Solution: Writing signs of /(a:), + + +, there is no change, hence no
positive root.
' Writing signs of /(—»;),— + + , there is one change, hence no more
than one negative root. Since there can be only one real root there must
be two imaginary roots.
(b) a;s + 1 = 0. (c) cc* - 2 = 0.
(d) a;8 - a; + 1 = 0. (e) a;^ - x + 1 = 0.
(f ) xi + x + l = 0. (g) a;5 + a;2 + 1 = 0.
(h) a;8-6a;2 + 4a;-l = 0. (i) x^-2x*'-3x'+ ix'+x + 1 = 0.
(j) x^ + 2xi - 6x' - ix" + X - 1 = 0.
190 ADVANCED ALGEBRA
178. Integral roots. In finding the rational roots of an equa-
tion we make use of the following
Theorem. If the equation
af+ajaf-i +••■+«„ = (? (1)
(where the a's are integers) has any rational root, such root must
be an integer.
Suppose - he a fraction reduced to its lowest terms which
satisfies the equation.
Tben |! + ^4-... + «„ =
is an identity.
Then clearing of fractions and transposing,
P" = - qiaip"-^ + ■■■ + «n2""0-
Thus some factor of y is a factor of ^", that is, of ^ (p. 52), which
P
contradicts the hypothesis that - is reduced to its lowest terms.
Thus all the rational roots of the equation are integers, which
as we know (§ 170) are factors of «„.
179. Rational roots. If we seek the rational roots of
aoX" H \-a„ = 0,
where ag -4=- 1, we can multiply the roots by a properly chosen
constant (§ 175) and obtain an equation of form (1) above whose
integral roots may easily be found by synthetic division.
BxAMPLE. What rational roots, if any, has
3a»+llx-14 = 0? (1)
Multiply the roots by 3, 3 a* + 99 x - 378 = 0.
Divide by 3, a' + 33 a; - 126 = 0. (2)
Since by Descartes' rule of signs equation (1) has only one positive root
and no negative root, we do not need to carry the table further than to test
for a positive root.
THEORY OF EQUATIONS 191
y Form a table of values for equation (2) by synthetic division.
37^ We need only to try the factors of 126 (§ 170).
Thus (2) has the root 3. Hence the original equation (1) has
the root 3 -=- 3 = 1.
62
EuLE. To find all the rational roots of an equation, trans-
form the equation so that the first coefficient is +1.
Find the maximum number of positive and negative roots by
Descartes^ rule of signs.
Find the integral roots of this equation by trial, and the
roots of the original equation by dividing the integral roots
found by the constant by which the roots were multiplied.
By th.e Theorem § 178 we are assured that all the rational
roots can be found in this way.
EXERCISES
Find all the rational roots of the following equations.
1. 4a8 = 27(a; + l). 2. 15a;8 + 13a;2- 2 = 0.
3. 4a;S-5ci;-6 = 0. 4. a*- 2f a;2+ 2|a; - 1 = 0.
5. 4a!8-8a;2-a; + 2 = 0. 6. 3a^ - 8a;S - 36a;2 + 25 = 0.
7. 4a;8-4a;2+z-6 = 0. 8. 3a;3 + 13a;2 + 11a; - 14 = 0.
9. 4a;3+16a;2-9x-36 = 0. 10. 2a;3 -21a;2 + 74a; - 85 = 0.
H. 6a;8-47a;2 + 71a; + 70 = 0. 12. 12a;3 - 52a;2+ 23a; + 42 = 0.
13. 6a;3-29a;2+ 53a;-45 = 0. 14. 6a;*- a;^- 8a;2 - 14a; + 12 = 0.
15. 27a;8 + 63a;2 + 30x-8 = 0. 16. 2a;*-13a;S + 16x2- 9x + 20 = 0.
17. 3x3 -26a;2+ 52a; -24 = 0. 18. 6x*- xS- 49x2+ 65x - 50 = 0.
19. 18x3+ 81x2+ 121x + 60 = 0. 20. 12x*+ Cx" - 24x2- 9x + 9 = 0.
21. lOx* + 18x' - 16x2 + 8x - 20 = 0.
22. 9x* + 15x8 - 143x2 +41X + 30 = 0.
23. 36x*-72xS-31x2+67x + 30 = 0.
24. 24x4-108xS + 324x2-240x + 60 = 0.
180. Diminishing the roots of an equation. In the preceding
sections we have solved completely the problem of finding the
rational roots of an equation. We now pass to the problem of
192 ADVANCED ALGEBRA
finding the approximate values of the irrational roots of an equa-
tion. In carrying out the process that we shall develop it is
desirable to form an equation whose roots are equal respectively
to the roots of the original equation each diminished by a constant.
Let f{x) = «„»" + aia^-i + ••• + «„ = 0, (1)
whose-roots are a-i, a^, ■■■, ar„. Let a be any constant. We seek
an equation whose roots are a^ — a, a^ — a, • ■ ■ , a„ — a.
If we let a stand for any one of the roots of (1), since /(«) =
(p. 33), we see that
/(a + a) = is satisfied by a: — a,
that is, /(or — a + «)=/(«) = 0.
Thus to form the desired equation replaxje x by z + a. We
obtain
f{x) = f{z + a)= ao(z + ay + a^(z + a)»-i + . . . + a„ = 0.
Developing each term by the binomial theorem and collecting
like powers of s, we get an equation of the form
f(x) = F(z) = Aoz- + A^z-^ + • • • + 4„ = 0, (2)
where the ^'s involve the a's and a. This is the equation desired.
We now seek a convenient method of finding the values of the
coefficients A^, A-i, A^, ■■■, A„ when Uq, %, a^, •■-, a„ are given
numerically. Now ^„ is the remainder from the division of F(z)
by s. But since F(z) =f(x) and z = x — a, the remainder from
dividing F{») by & is id.entical with the remainder from dividing
/(x) hj X — a. Thus ^„ is the remainder from dividing f(x) by
X — a. Furthermore, since A„_^ is the remainder from dividing
F(z) A fix") A
—^ ~ by s, it is also the remainder from dividing "^-^^^^
s X — a
hj X — a. The process may be continued for finding the other A's.
We may then diminish the roots of an equation by a as follows :
EuLE. The constant term of the new equation is the remaiib-
der from dividing f(x) hy x — a.
THEORY OF EQUATIONS
193
The coefficient of z in the new equation is the remainder from
dividing the quotient just obtained hy x — a.
The coefficients of the higher powers of z are the remainders
from dividing the successive quotients oitained hy x — a.
Example. Form the equation whose roots are 2 less than the roots of
st-2a;«-4a;2 + x-l = 0.
The diyisions required by the rule we carry out synthetically (p. 169).
1-2 -4 +1 - 1[2
+ 2 +0 -8 -14
1+0-4-^7
+ 2 +4 +0
-15
1 + 2 +0
+ 2 +8
-7
1 + 4
+ 2
+ 8
1+6
The desired equation is
x4 + 6!i;s-
f 8a;2-7a;-l
i5 = 0.
181. Graphical interpretation of decreasing roots. If an equa^
tion has roots a units less than those of another equation, if a is
positive its intersections with the X axis or with any line parallel
to the X axis are a units to the left of the corresponding inter-
sections of the first equation. It is, in fact, the same curve, except-
ing that the Y axis is moved a units to the right. If a is negative,
the Y axis is moved to the left.
EXERCISES
Plot, decrease the roots by a units, and plot the new axes.
1. a* - 3»3 - 2x - 3 = 0. a = 3.
Solution: 1-3- 0- 2-3[3
(1)
+ 3
-
-
-6
1_0 -0-2
+ 3 + 9+27
-9
1 + 3 + 9
+ 3 +18
+ 25
1 + 6
+ 3
+ 27
1 + 9
194
ADVANCED ALGEBRA
1
Y
1
T-
\
\
\
\
V
\
,0
0'
*S
V
r
\
/
>
V
\
y
J
\
/
\
\
/
Thus the required equation is
x*+9a;8 + 27a;2 + 25a; -9 = 0.
(2)
X
y
- 3
1
- 7
2
-15
3
- 9
- 1
+ 3
In the figure one square on the 7" axis repre-
sents two units of y, and two squares on the
X axis represent one unit of x.
2. a;*-16 = 0. a = 2.
4. x*-%x^ + l = 0. a = .2.
6. a:3-4a;2-2 = 0. a = .5.
8. x^ + ix^ + x-& =,0. a = — .4.
10. a;8 - 3a;2 + X - 1 = 0. a = - .3.
a = - 2.
3. x3 - 8 = 0. o = 1.4.
5. a;S + 4a._8 = o. o = 3.
7. xs + 2x + 5 = 0. a = -l.
9. a!»-2a;2-|-8x-7=0. a = 2.
11. x*-3a;2 + 2a;-2 = 0.
12. x8 - 15V + 7a; + 125 = 0. a = 5.
13. 2xS-6x2 + 4x-3 = 0. a = -3.
14. X* + 6x8 + xox2 + x-l = 0. a = -l.
182. Location principle. If when plotting an equation y =f(x)
the value x = a gives the corresponding value of y positive and
equal to c, while the value x = h gives the corresponding value
of y negative, say equal to — d, then the y .
point on the curve a; = a, y = c is above
the X axis, and the point on the curve
x = h,y = — d\s below the X axis. If our
curve is unbroken, it must then cross the
X axis at least once between the values
K = a and x = h, and hence the equation
must have a root between those values
of X. The shorter we can determine this
interval a to S the more accurately we can iind the root of the
equation. This property of unbrokenness or continuity of the
graph of 2/ = UaX" + aix"~^ + 1- «„ we assume. We assume
then the following
THEORY OF EQUATIONS
195
Location principle. When for two real unequal values of x,
say x= a and x = h, the value of y =f(x) has opposite signs,
the equation f(x) = has a real root between a and h.
Illustkation. The equation f(x) =a;' + 3a; — 6 = has a
root between 1 and 2. Since /(1) = -1, /(2) = 9.
183. Approximate calculation of roots by Horner's method.
We are now in a position to compute to any required degree of
accuracy the real roots of an equation. Consider for example
the equation ^s + 3^_20 = 0. (1)
Form the table of values for plotting the equation
x' + 3x — 20 = y.
By the location theorem we find
that a root is between + 2 and + 3.
To find more precisely the position
of the root we might estimate from
the graph the position of the root
and substitute say 2.3, 2.4, and so
on, until we found two values be-
tween which the root lies. We can
gain the same ' result with much
less computation if we first dimin-
ish the roots of the equation so that
the origin is at the less of the two integral values between which
we know the root lies. Here we decrease the roots of (1) by 2,
1 + + 3 - 20|2
+ 2 + 4-1-14
- 6
-20
-16
- 6
+ 16
-1
-24
1 + 2+ 7
+ 2+8
1 + 4
+ 2
+ 15
1+6
The equation whose roots are decreased by 2 is
a;' + 6x'' + 15a;-6 = 0.
(2)
196 ADVANCED ALGEBRA
We know that (2) lias a root between and 1, since equation (1)
has a root between 2 and 3. From the graph we can estimate the
position of the root. Having made an estimate, say .3, it is neces-
sary to verify the estimate and determine by synthetic division
precisely between which tenths the root lies. Thus, trying .3, we
°^*^"^ 1 + 6.0 + 15.00 - 6.000[^
+ 0.3+ 1.89 + 5.067
1 + 6.3 + 16.89 - 0.933
which shows that for x = .3 the curve is below the X axis, hence
the root is greater than .3. But we are not justified in assuming
that the root is between .3 and .4 until we have substituted .4
for X. This we proceed to do.
1 + 6.0 + 15.00 - 6.000 [^
+ 0.4+ 2.56 + 7.024
1 + 6.4 + 17.66 + 1.024
Since the value of y is positive for x = .4, the location principle
shows tha,t (2) has a root between .3 and .4, that is, (1) has a root
between 2.3 and 2.4.
To find the root correct to two decimal places, move the origin up
to the lesser of the two numbers between which the root is now
known to lie. The new equation will have a root between and .1.
This' process is performed as follows :
' 1 + 6.0 + 15.00 - 6.0001^
+ 0.3 + 1.89 + 5.067
1 + 6.3 + 16.89
+ 0.3 + 1.98
- 0.933
1 + 6.6
+ 0.3
+ 18.87
1 + 6.9
Thus the new equation is
x" + 6.9 w^ + 18.87 X - .933 = 0. (3)
This equation has a root between and .1. We can find an
approximate value of the hundredths place of the root by solving
THEORY OP EQUATIONS 197
the linear equation 18.87 x — .933 = 0, obtained from (3) by drop-
ping all but the term in x and the constant term.
QOO
^^'^^ ^ = 18:87 = •«*■
This suggestion must be verified by synthetic division to deter-
mine between what hundredths a root of (3) actually lies.
1 + 6.90 + 18.870 - 0.9330|.04
-f 0.04 -I- 0.277 -t- 0.7658
6.94 + 19.147 - 0.1672
Thus the curve is below the X axis at x = .04 and hence the
root is greater than .04. We must not assume that the root is
between .04 and .05 without determining that the curve is above
the X axis at a; = .06.
1 -I- 6.90 -I- 18.870 - 0.9330|.05
+ 0.05+ 0.347 + 0.9608
6.96 + 19.217 + 0.0278
Thus the curve is above the X axis at a; = .05. By the location
principle (3) has a root between .04 and .05, that is, (1) has a
root between 2.34 and 2.35. We say that the root 2.34 is correct
to two decimal places. If a greater degree of precision is desired,
the process may be continued and the root found correct to any
required number of decimal places.
The foregoing process affords the following
EuLE. Flot the, equation. Apply the location principle to deter-
mine hetween what consecutive positive integral values a root lies.
Decrease the roots of the equation ly the lesser of the two
integral values between which the root lies.
Estimate from the plot the nearest tenth to which the root of
the new equation lies, and determine hy synthetic division pre-
cisely the successive tenths hetween which the root lies.
Decrease the roots of this equation hy the lesser of the two tenths
hetween which the root lies, and estimate the root to the nearest
hundredth hy solving the last two terms as a linear equation.
198
ADVANCED ALGEBKA
Determine hy synthetic division precisely the hundredths inter-
val in which the root must lie.
Proceed similarly to find the root correct to as many places as
may be desired.
The sum of the integral, tenths, and hundredths values next less
than the root in the various processes is the approximate value
of the root.
To find the negative roots of an equation f(x) = 0, find the
positive roots of /(— x) = and change their signs.
When all the roots are real a check to the accuracy of the com-
putation may be found by adding the roots together. The result
should be the coef&eient of the second term.
EXERCISES
Find the values of the real roots of the following equations correct to
three decimal places.
1. a;3+4a;2 + a; + l = 0. (1)
Solution : Since by Descartes'
rule of signs there are no posi-
tive roots, we form the equation
/(— k) = and seek its positive
root.
Thus
x'-4:X^ + x-l = 0. (2)
Plot the equation (2) set equal
to y. In the figure two squares
are taken as the unit of a;.
There is a root of this equa-
tion between 3 and 4.
Decrease the roots of (2) by 3,
1-4 -1-1 -1[3
+ 3-3-6
Y)
I
X
k "
/
V
1
,
N
2
1
3
1
j
4
1
V
J
\
1
1
^
>
I
■ 1
3
-7
•7
+ 3
1-1 -2
-1-3 -1-6
1 + 2
+ 3
1 + 6
+ 4
THEORY OF EQUATIONS 199
The equation is x' + 6x^ + ix — 1 = 0. (3)
From the plot we estimate the root of (3) at .8.
Verify, l + 5.0 + 4.00 - 7.000 [S
+ 0.8 + 4.64 + 6.912
+ 5.8 + 8.64 - 0.088
1 +5.0 + 4.00- 7.000 LJ
+ 0.9 + 6.31 + 8.379
+ 5.9 + 9.81 + 1.379
Thus the root is determined between .8 and .9.
Decrease the roots of (3) by .8,
1+5.0 + 4.00 - 7.000 [^
+ 0.8 + 4.64 + 6.912
1+5.8 + 8.64
+ 0.8 + 5.28
- .088
1 + 6.6
+ 0.8
+ 13.98
The equation Is
1+7.4
x' + 7.4a;2 + 13.92a; - .088 = 0.
(4)
088
Estimate the root of (4) at a; = = .006.
^ ' 13.92
Verify, 1 + 7.400 + 13. 920000 - .088000 1.006
+ 0.006 + 0Q.'044436 + .083784
+ 7.406 + 13.964 - .004216
1 + 7 .400 + 13.920000 - .088000 1.007
+0 .007 + 00.051849 + .097804
+ 7.407 + 13.972 + .009804
Thus the root of the equation (1) correct to three decimal places is - 3.80li
2. a;s - 4 = 0. 3. a;* - 3 = 0.
4. x^ + x = 20. 5. Sx^ - 5a;3 = 31.
6. x' + x^ = 100. 7. a;3 - a; - 33 = 0.
8. a;* + a; - 100 = 0. 9. a;S - 8a; - 24 = 0.
10. x* - 4a;8 + 12 = 0. 11. x* + a;^ + x = 111.
12. x8 _ a;2 + a; - 44 = 0. 13. x" + 10a; - 13 = 0.
14. a;» + 3a;2 - 2a; - 1 = 0. 15. x^ + a;2 ^. j; _ 99 = 0.
16. a;8 - 9x2 - 2x + joi = 0. 17. x* + x^ + x^ - 88 = 0.
18. X* - 12x8 - 16x + 41 = 0. • 19. x8 - 6x2 + 5a; + 11 = 0.
20. x8-l6x2 + 35x+ 50 = 0. 21. 2x* - 4x3 + 3x2 - 1 = 0.
22. ,3x* - 2x3 - 21x2 - 4x + 11 = 0. 23. Ox^ - 45x2 + 34x + 37 = 0.
200
ADVANCED ALGEBRA
184. Roots nearly equal. Suppose we wish to find tlie positiTe
roots, if any exist, of
a;= + 17a;2- 46a; + 29 = 0. (1)
By Descartes' rule of signs we see that there can be
only two positive roots. We obtain the adjacent table
of values. From the plot that these values indicate we
cannot tell whether any real root exists between 1 and 2,
but if it does exist the plot indicates that it is nearer
1 than '2.
Decrease the roots of (1) by 1,
1 + 17 - 46 + 29[1
+ 1+18-28
y
+ 29
+ 1
+ 13
+ 71
1 + 18 - 28
+ 1+19
+ 1
1 + 19
+ 1
- 9
1+20
The new equation is
a;« + 20a;2-
-9a; + l =
(2)
Estimate the root of (2) at .2 and carry the origin up to .2
and also up to .3.
1 + 20.0 - 9.00 + LOOOL^
'+ 0.2 +4.04 -0.992
1 + 20.0 - 9.00 + 1.000[^
+ 0.3 + 6.09 - 0.873
1 + 20.2 -4.96
+ 0.2 +4.08
1 + 20.4
+ 0.2
+ 0.008
-0.88
1 + 20.3 - 2.91
+ 0.3 + 6.18
+ 0.127
1 + 20.6
+ 0.3
+ 3.27
1 + 20.6
1 + 20.9
By Descartes' rule of signs on the numbers obtained by moving
the origin to .3, it is seen that there are no positive roots of (2)
greater than .3, while the rule would indicate that there might
be roots greater than .2. We consider the equation
;«;» + 20,6 a;^ - 0.88 a; + 0.008 = 0. (3)
THEOKY OF EQUATIONS 201
: .009/
Estimate the roots of (3) at
_ 0.008
"'"O.SS '
Verify, 1 + 20.60 - 0.880 + 0.00800[.01
+ 0.01 + 0.206 - 0.00674
1 + 20.61 - 0.674 + 0.00126
1 + 20.60 - 0.880 + 0.00800|.02
+ 0.02 + 0.412-0.00936
1 + 20.62 - 0.468 - 0.00136
This determines a root of (3) between .01 and .02.
1 4- 20.60 - 0.880 + 0.00800|.03
+ 0.03 + 0.619-0.00483
1 + 20.63 - 0.161 + 0.00317
This determines another root of (3) between .02 and .03.
Decrease the roots of (3) by .01,
1 + 20.60 - 0.880 + 0.00800|.01
+ 0.01 + 0.206 - 0.00674
1 + 20.61 - 0.674
+ 0.01 +0.206
1 + 20.62
+ 0.01
0.00126
0.468
1 + 20.63
The new equation is
x^ + 20.63 x^ - 0.468 x + 0.00126 = 0. (4]
Estimate the root of (4) at a; = " = .002.*
0.4oo
Verify, 1 + 20.630 - 0.468 + 0.00126|.002
+ 0.002 + 0.041 - 0.00085
1 + 20.632 - 0.427 + 0.00041
1 + 20.630 - 0.468 + 0.00126|.003
+ 0.003 + 0.062-0.00122
1 + 20.633 - 0.406 + 0.00004
* "We observe that in these two cases the estimated values of the roots are shown hy
the verification to he inaccurate. This should insure great care in making the verifi*
CAtion* Tk0 Qstvvmt^d vc^lu^s shov/ld never be asswned to be c^ccwate without veriJicatUm
202 ADVANCED ALGEBRA
This indicates that a root is between .003 and .004.
Verify, 1 + 20.630 - 0.468 + 0.00126]. 004
+ 0.004 + 0.083-0.00154
1 + 20.634 - 0.385 - 0.00028
This determines a root of (3) between .003 and .004.
Thus one root of (1) correct to three decimal places is 1.213.
The other root could be found similarly to be 1.229.
EXERCISES
Find all the real roots of the following equations correct to three decimal
places.
1. a;S-7x + 7 = 0.
2. 7a;3- 8x2 -14a; + 16 = 0.
3. 2x5 _ 4x8-3x2 + 6 = 0.
4. 4x*- 5x3 -8x + 10 = 0.
5. 3x' - 10x2 -33x + 110 = 0.
CHAPTEE XVIII
DETERMINANTS
185. Solution of two linear equations. We have abeady
treated the solution of linear equations in tvo variables and
stated (p. 47) the method of solving three or more linear equa^
tions in three or more variables. This latter process is rather
laborious and can be very much abridged and also developed
more symmetrically by the considerations of the present chapter.
Let us solve the equations
u.iX + b^y = Ci, (1)
a^x + \y = Cj. (2)
Multiply (1) by Sj ^^^ (2) by 6i, and we obtain
a-J)^x + h^h^y — b^c^
a^biX + bib^y = SiCg
Subtracting, we get (aib^ — a^bi) x = b^c^ — b^o^, ^
or if a,b, - a,b, ^0, x = h^^zh^. (3)
Similarly, we obtain y = ^7^ ~- C4')
We note that the denominators of the expressions for x and y
are the same. This denominator we will denote symbolically by
the following notation :
01162 — ctaSi =
(5)
The symbol in the right-hand member is called a determinant.
Since there are two rows and two columns, this determinant is
said to be of the second order. The left-hand member of the equa-
tion is called the development of the determinant. The symbols
"xj ^1) <^2) &2 are called elements of the determinant, while the ele-
ments ax and b^ are said to comprise its principal diagonal.
203
204
ADVANCED ALGEBRA
EuLE. The development of any determinant of the second
order is obtained ly subtracting from the product of the ele-
ments on the principal diagonal the product of the elements on
the other diagonal.
ThTish ^ l = a;yi-a!iy; I! fl = 10-l2=-2.
m yi\ |4 5|
Evidently each term of the deTelopment contains one and only
one element of each row and each column, that is, for instance,
the letter a and the subscript 1 appear in each term of (5) once
and only once.
We can now rewrite the solution (3) and (4) of equations (1)
and (2) in determinant form :
(6)
Cl
h
Ol
Cl
Oi
h
; y =
aa
Cj
«!
h
«!
hx
aj
h
ttj
h
It is noted that the numerator of the expression for x is farmed
from the denominator by replacing the column which contains
the coefficients of x, a^ and a^, by the constant terms Ci and c^.
Similarly, in the numerator of the expression for y, 6i and Jj of
the denominator are replaced by Ci and e^.
One should keep in mind that a determinant is merely a sym-
bolic form of expression for its development. In the case of
determinants of the second order the introduction of the new
notation is hardly necessary, as the development itself is simple ;
just as we should scarcely need to introduce the exponential
notation if we had to consider only the squares of numbers. It
turns out, however, as we shall see, that we are able to denote by
determinants with more than two rows and columns expressions
with whose development it would be very laborious to deal.
186. Solution of three linear equations. Let us solve the
equations
Uix + biy + ci« = di,
a^x + b^y + CaS = d^,
asX + h^y + Cs» = d^.
(1)
(2)
(3)
DETERMINANTS
205
Eliminating y from (1) and (2) and (1) and (3), we obtain
(ttiJa — a^h^x + (S2C1 — JiCj)*! = ^162 — djkx,
(a,bi — a^bt)x + (cjJi — S,Ci)s = rfa*i — <^i*8-
Eliminating «,
[(aiJjj — flSa^i) ("s^i — hoi) — (aah — aA) (*2«i — ^162)] a:
= (dih — <^2*i) («8*i — *8«i) — («^8*i — <^i*a) (^aCi — iiCj),
Developing, canceling, and dividing by 61, we obtain
dib^Cs + d^Ci + dsbiC^ — dji^c^ — dgb^Oi — d^b^es
aib^Ci + "zSa*! + »8^iC2 — aibsC^ — asb^Ci — ajJiCs
Eollo-wing the analogy of the last section, we write the denomi-
nator
(4)
c('ib2Cs + ajfta"! + as^iCa — aJiaC^ — a^o^ — a^b^c^
as
b,
6a
.(5)
The right-hand member of this equation we call a determinant
of the third order, and the left-hand member its development.
As in the determinant of the second order, the elements %, b^, c^
comprise the principal diagonal; each term of the development
contains one and only one element of each row and each column,
and all possible terms so constructed are included in the develop-
ment. The signs of the terms of the determinant of the third
order may be kept in mind by the following device.
Eewrite the first and second columns to the right of the deter-
miaant as follows :
a, 5i ci
<h h
ffij ^2 Cj
£52 62
flSa 63 Cs
as bs
The positive terms are found on the diagonals running down
from left to right, the negative terms on the diagonals running '
up from left to right.
The numerator of the fraction (4) expressed in determinant
notation is
61
206
ADVANCED ALGEBRA
d.
\
Cl
d.
h
"2
d.
h
Cs
o-i
h
Cl
a^
h
Cs
as
h
Cs
(h
^i
Cl
a^
rf.
Ca
as
C^s
Cs
«!
*i
Cl
aj.
b.
C2
as
h
Cs
a,
h
d.
ffla
h
d.
as
h>
d.
«!
h
Cl
as
h
"2
as
b.
Cs
We can find similarly the values of y and « that satisfy equa-
tions (1), (2), and (3). The solutions of the equations in determi-
nant form are as follows :
(6)
The same principle observed on p. 204 for forming the deter-
minants in the numerators of the expressions for x and y may be
followed here. The determinant in the numerator of the expres-
sion for X, y, or s is found from the denominator by replacing the
column that contains the coefB.cients of the variable in question
by a column consisting of constant terms. Thus in the numerator
of s -we find the column tZj, d^, dg replacing the column c^, Cj, Cj of
the denominator,
EXERCISES
1. Find the value of the following determinants.
= 18 + 4 + 0-6-8-0 = 8.
3 2 1
(a)
4 6 2
1 1
•
3 2 1
Solution :
4 6 2
10 1
4
3
1
(b)
1
2
6
1
1
3
4
1
(d)
2
4
1
6
2
1
(c)
2
1
1
2
2
1
1
(e)
6
3
3
4
2
3
(f)
Ci)
a X y
6 c
c 6
c 6
-c a
-& -0
o|
(g)
(i)
c
b
c
a
6
a
DETERMINANTS
207
2. Solve the following equations by determinants.
2x + 3y = i,
(a)
x-2y = l.
Solution ; Using the expressions (6), p. 206, we obtain
Check: -V-
4 3
2
4
1 -2
- 8 - 3 11
" -4-3" 7 ' ^~
1
1
2-4 2
2 3
2
3
-4-3 7
1 -2
1
-2
--f =
1 = 1.
w^!l«!::'(e)
'.x + Ty = l, x + 4y = 2,
■ 9y = 2. ^ ' 2x-3y = -l.
(e)
x-ly = 12,
1x + y = ll.
Tx + 6y = 2. ^ ' 3x~9y = 2. ^ ' 2x-Zy = -
3. Solve the following equations by determinants.
a; + 2/ + z = 2,
(a) 1 + 32^-4 = 0,
2/ - 2 2 = 6.
Solution : Rearranging so that terms in the same variable are in a column,
and supplying the zero coefficients, we get
x+ y + z = 2,
x + Zy + (lz = i,
Ox+ y-2z = Q.
By (6), p. 206, a: =
2
1
1
4
3
6
1
-2
1
1
3
1
-2
2
1
4
6
-2
1
1
3
1
-2
-12+0 + 4-18 + 8 + 0_-18
-6+0+1-0-0+2 ~ -Z'
■8 + + 6-0 + 4 +
-3
z =
1 2
3 4
1 6
18 + + 2-0-6-4 10
111 -3
13
1-2
Check : 6 + (- f) + (- Y) = 6 - if- = 6-4 = 2.
10
3'
208 ADVANCED ALGEBRA
2x + Sy = 12, lx-ly = 0,
(b)3a; + 2z = lL (c) |a; - |z = 1,
32/ + 4z = 10. ^z-iy = 2.
x + y-z=:n, 2x + 2y = T,
{A) x + z'-y = 13, (e)7x+ 92 = 29,
y + z- x = 7. y + 8z = n.
x + y + z = 100, x + y + 2z = 3i,
(f)3a;-2z = 4, (g) a; + 2y + z = 33,
5s^-4z = 0. 2x + y + z = 32.
x + y + z = 9, ax + by — cz = 2db,
(h) x + 2y + Sz = li, (i) by + ez-ax=2 be,
a: + 3 y + 6 z = 20. cz + ax — by = 2ac.
.2x + .3y + Az = 29, Zx + 2y + 3z = n0,
(j) .3a; + .4y + .5z = 38, (k) 6x + y-4:Z = 0,
.ix + .5y + .ez = 5l. 2x-3y + z = 0.
187. Inversion. In order to find the development of determi-
nants with more than three rows and columns, the idea of an inver-
sion is necessary. If in a series of positive integers a greater
integer precedes a less, there is said to be an inversion. Thus in
the series 12 3 4 there is no inversion, but in the series 12 4 3
there is one inversion, since 4 precedes 3. In 1 4 2 3 there are two
inversions, as 4 precedes both 2 and 3 ; while in 1 4 3 2 there are
three inversions, since 4 precedes 2 and 3, and also 3 precedes 2.
188. Development of the determinant. In the development of
the determinant of order three we have
= ajb^Cs + a^bgCi + as&iCj — as^aCi — chh^s " ^i^sCa- (1)
«! bi Cj
"s ba Cs
If we keep the order of letters in each term the same as their
order in the principal diagonal (as we have done in the development
above), it is observed that the subscripts in the various terms
take on all possible permutations of the three digits 1, 2, and 3.
The permutations that occur in the positive terms are 12 3,
2 3 1, 3 1 2, in which occur respectively 0, 2, and 2 inversions.
The permutations that occur in the subscripts of the negative
terms are '3 2 1, 2 1 3, 1 3 2, in which occur respectively 3, 1,
and 1 inversions.
DETEKMINANTS 209
Thus in the subscripts of the positive terms an even numher
of inversions occur, while in the subscripts of the negative terms
an odd number of inversions occur. This means of determining
the sign of a term of the development we shall assume in general.
When we have a determinant with n rows and columns it is
called a determinant of the »th order. The development of such a
determinant is defined by the following
EuLE. The development of a determinant of the nth order is
equal to the algebraic sum of the terms consisting of letters fol-
lowing each other in the same order in which they are found in
the principal diagonal hut in which the subscripts take on all
possible permutations. A term has the positive or the negative
sign according as there is an even or an odd number of inver-
sions in the subscripts.
This means of finding the development of a determiaant is use-
ful in practice only when the elements of the determinant are
letters with subscripts such as in (2) below. When the elements
are numbers we shall find the value of the determinant by a more
convenient method.
In this statement it is assumed that the number of inversions
in the subscripts of the principal diagonal is zero. If this number
of inversions is not zero, the sign of any term is + or — accord-
ing as the number of inversions in its subscripts differs from the
number in the subscripts of the principal diagonal by an even or
an odd number.
Since each term contains every letter a, b, ■ ■■, k and also every
index 1, 2, • • •, n, one element of each row and column occurs in
each term.
In the determinant of the fourth order
(2)
the terms a^itCidi and at&sCgcZi, for instance, have the minus sign, as 2 4 1 3 has
three inversions and 4 2 3 1 has five inversions; while the terms aibiC^ds and
o^dgCj^x have two and six inversions respectively and hence have the positive sign.
«!
5i
"1
di
02
62
C2
*2
as
h
cs
ds
Oi
bi
Bi
di
210
ADVANCED ALGEBRA
189. Number of terms. We apply the theorem of permutar
tions to prove the following
Theorem. A determinant of the nth order has n ! terms in
its development.
Since the' number of terms is the same as the number of per-
mutations of the n indices taken all at a time, the theorem follows
immediately from the corollary on p. 145.
190. Development by minors. In the development of the
determinant of order three, p. 208, we may combine the terms as
follows :
ai &i Ci
Ctg Og ^2
^8 ^S ^8
= % (SjCa — *s«i!) — «2 (p\Os — ^sCi) + % (*iCa — ho^
6, c^
(1)
We observe that the coefB-cient of a^ is the determinant that
we obtain by erasing the row and column in which Oj lies. A
similar fact holds for the coefScients of a^ and a^. The determi-
nant obtained by erasing the row and column in which a given
b
element lies is called the minor of that element. Thus / 'is
the minor of a^. We notice that in the above development by
minors (1) the sign of a given term is -t- or — according. as the
sum of the number of the row and the number of the column
of the element in that term is even or odd. Thus in the first
term % is in the first row and the first column, and since 1 -|- 1 = 2,
the statement just made is verified for that ease. Similarly, ajj is
in the first column and the second row, and since 1 -|- 2 = 3 is
odd, the sign is minus and the law holds here. The last term
is positive, which we should expect since 0.3 is in the first column
and the third row, and 1-1-3 = 4. The proof for the general
validity of this law of signs is found on p. 215.
The elements of any other row or column than the first may
be taken and the development given in terms of the minors with
DETERMINANTS
211
«1
"2
= — fflj
Cl
+ h
«1
Cl
— ^2
a-i
Si
Cs
as
Cs
as
h
«8
respect to such elements. For instance, take the development
with, respect to the elements of the second row,
«i h
as is
The rule of signs is the same as given above; that is, for
instance, the last term is negative, as Cj is in the third column
and the second row, and 2 + 3 = 5. By generalizing these con-
siderations we may find the development of a determinant by
minors by the following
EuLE. Write in succession the elements of any row or column,
each multiplied hy its minor.
Give each term a -\- or a — sign according as the sum of the
number of the row and the nuwiher of the column of the element
in that term is even or odd.
Develop the determinant in each term hy a similar process
until the value of the development can he determined directly hy
multiplication.
That this rule for development gives the same result as the definition given in
§ 188 we have seen for a determinant of order three. The fact holds in general, as
we shall prove (p. 215) .
EXERCISES
1. In the determinant of order four on p. 209 what sign should be pre-
fixed to the following terms ?
(a) Cibsa^di;
Solution: Cibsa^di = a2bsc^dl. lu 2 3 4 1 there are three inversions. The
sign should be minus.
(b) atbiCadi. (c) OnbiCids- (d) 6401^302.
(e) dsbittiCi. (f) d^aiCib-i. (g) Ciasdibt.
2. Develop by minors the following and find the value of the determinant
3 2
4
(a)
2 1 4
3 16
Solu
tlon:
3
3
3
2 4
1 4
1 6
= ^1 6
2^ * -^3^ *
-^ 1 6+^1 4
= 3(6 - 4)- 2(12 - 4) 4- 3(8 - 4) = 6 - 16 -1- 12 = 2.
212
ADVANCED ALGEBRA
1
4
6
(b)
7
8
2
1
3
1
a
6
(e)
d
c
e
/
(c)
(f)
(h)
2 1
2 1
3 4
a 6
a 6
a &
2 3 1
3 1
4 2 12
5 7
21
(d)
3 7 1
-2 3-1
6
(g)
c
d e
•
1
1 2
1 2
Solution : Develop with respect to the elements of the first column,
2 3 11
3 11
112
2 12
12 3
-3
3 11
2 12
12 3
+ 4
3 11
112
12 3
-('15 SI-15 ah'l; ^l)-K=|2
+4
2
3| "|2 3
= 2(-l + 2 + 0)-3(-3-
= 2 + 12 -12- 3 = -1.
11^ ^l+il^ ^\ s/sP
IL J + l|j 2|rT|l
-2
-1
3 11
1 1 2
2 12
1 1
2 3
1 1
1 2
+ 1
+2
1 1
1 2
1 1
1 2
i + l) + 4(-3-l + l)-3 (0-1+2)
(i)
2
6
3
9
8
1
3
6
1
8
2
1
4
(i)
12 3 4
1111
3 11
3 3 3
Hint. It is always advisable to develop with respect to the row or column that
has a maximum number of elements equal to zero,
a 6
6 a
6 a ■
(k)
(m)
(0)
(q)
(1)
6
a
2
3
4
1
2
3
3
6
2
3
1
1
2
3
2
1
X
a
6
b
X
a
a
b
X
X
y
y
X
V
X
y
X
y
a
CLl &i Ci dx
hz C2 dz
(n)
(P)
W
Cs (is
d4
ai
02 62
03 63 Cs
04 bi c^ di
a g
e a f
o
d c b a
X a b c
c X a b
b c X a
a b c X
DETERMINANTS
213
191. Multiplication by a constant. In this and the following
sections we shaE give a number of theorems on determinants
which greatly facilitate their evaluation and which make a proof
for the solution in terms of determinants of any number of linear
equations in the same number of variables a simple matter.
Theorem. If every element of a row or a column is multi-
plied hy a number m, the determinant is multiplied by m.
Suppose that every element of the first row of a determinant
is multiplied by m. Since each tprm of the development contains
one and only one element from the first row, every term is multi-
plied by m, that is, the determinant is multiplied by m.
Illustkation.
= m,aib^Cf + ma^bsCi + magbic^ — ma^h^c^ — Tna^biC^ — ma^b^Ci
ai a
! »8
= m
bi 62 bs
Ci C2 C3
•
6 4 1
Similarly,
8 3 2
=
10
4 1
2-3
2-4
2-5
= 2
192. Interchange of rows and columns. We now prove the
Theorem. The value of a determinant is not changed if the
columns and rows are interchanged.
Take for instance the determinant of order four.
«1
bi
Cl
d.
<H
%
a.
ai
ffl2
b.
C2
d^
b.
b.
bs
b.
as
&s
C3
d.
Ol
C2
"s
Ci
ai
64
C4
d.
d.
d.
d.
d.
In each of these determinants the principal diagonal is the same,
and hence the developments derived according to the statement on
214
ADVANCED ALGEBRA
p. 209 will be the same for e'acli determinant, since tlie terms ■will
be identical except for their order. The same reasoning is valid
for any determiaant.
193. Interchange of rows or columns. We now prove the
Theorem. If two columns or two rows are interchanged, the
sign of the determinant is changed.
Again let us take for example the determinant of order four and
fix our attention on the first and second rows. We must prove that
«!
h
<h.
d.
tta
h
C2
d.
ttj
h
62
d.
«i
h
«i
d.
as
bs
C3
dg
as
h
Cs
ds
at
h
Oi
d.
a*
b.
C4
d.
In the first determinant the principal diagonal is aib^Csdi, while
in the second the priacipal diagonal, a^bxCsdi, is obtaiaed from the
principal diagonal of the first determinant by one iaversion of
subscripts. Hence this term is foimd among the negative terms
of the first determinant.
Since the only difference between the second determinant and
the first is the interchange of the subscripts 1 and 2, evidently
any term of the second is obtained from some term of the first
by a single inversion. Thus if a single inversion is carried out in
every term of the first determinant, we obtain the various terms
of the second. But since this process changes the sign of each
term of the first determinant (p. 213), we see that the second
determinant equals the negative of the first. Similar reasoning
may be applied to the interchange of any two consecutive rows
or columns of any determinant.
Consider now the effect of interchanging any two rows which
are separated we will say by Te, intermediate rows. To bring the
lower of the two rows in question to a position next below the
upper one by successive interchanges of adjacent rows, we must
make k such interchanges. To bring similarly the upper of the two
rows to the position previously occupied by the other requires ifc + 1
further interchanges of adjacent rows. Hence the iuterchange of
the two rows is equivalent to 2 k +1 iaterchanges of adjacent
DETERMINANTS 215
rows, the effect of which is to change the sign of the determinant,
since 2 k +lis always an odd number.
194. Identical rows or columns. This leads to the important
Theorem. If a. determinant has two rows or two columns
identical, its valice is zero.
Suppose that the first and the second row of a determinant are
identical. Suppose that the value of the determinant is the num-
ber D. By § 193, if we interchange the first and second rows the
value of the resulting determinant is — D. But since an inter-
change of two identical rows does not change the determinant at
all, we have D = — D
or 2 D = 0, that is, D = 0.
OoEOLLAKY. If any row (or column) is m times any other
row (or column), the value of the determinant is zero. '
By § 191, the determinant may be considered as the product of
m and a determinant which has two rows (or columns) identical.
Hence this product equals zero.
195. Proof for development by minors. On referring to the
rule on p. 211 we observe that in order to show that the develop-
ment by minors is the same as the development obtained by the
definition on p. 209 we must prove the two following statements.
First. The coefficient of any element in the development of a
determinant (apart from sign) is the minor of that element.
Second. Uach element times its minor should have a + or
a — sign according as the sum of the number of the row and the
column of the element is even or odd.
Consider the element a^.
First. Each term that contains a-i must contain every other
letter than a, and the indices of these letters must take on all
permutations of the numbers 2, 3, • • • , to. This coefB.cient of a^
contains then by definition (p. 210) all the terms of its minor.
Second. Since in each term ai is in the first place, the only inver-
sions in the subscripts are those among the numbers 2, 3, • • • , ?i.
216
ADVANCED ALGEBRA
Hence the sign of each term in the coefficient of a^ is positive or
negative according as there is an even or odd number of inversions
in its suhscripts. Hence our theorem is established for the ele-
ment Oi.
Consider now any element, as d^, which occurs in the fifth row
and the fourth column. Interchange adjacent rows and columns
until ds is brought into the leading position in the principal
diagonal. This requires in all seven interchanges, three to get
the ds in the first column, and then four to get it into the first
row. This changes the sign of the determinant seven times,
leaving it the negative of its original value. By the reasoning
just given in the case of % the eoe£B.cient of d^ (which is now in
the position previously occupied by Ui) in the original determi-
nant would be the minor of d^, except that the signs would
all be changed. Hence the term consisting of d^ times its minor
has the — sign, and the theorem is proved for this case.
In general, to bring a term in the ith row and kth column to
the leading position requires i — l + k — l = i + k — 2 inter-
changes of adjacent rows or columns. This involves i + k — 2,
of what amounts to the same thing, i 4- k changes of sign. Hence
a positive or a negative sign should be given to an element times
its minor according as i -|- A; is even or odd.
196. Sum of determinants. We now prove the fact that under
certain conditions the sum of two determinants may be written in
determinant form. The fact that the product of two determinants
is always a determinant is extremely important for certain more
advanced topics in mathematics, but the proof lies beyond the
scope of this chapter.
Theorem. If each of the elements of any row or any column con-
sists of the sum of two numbers, the determinant may be written
as the sum of two determinants.
Por example, we have to prove that
ai + a\
h
Cl
«i
h
Cl
a\
*i
Cl
a^ + a'j
h
Ca
=
a^
h
"2
+
a'.
h
Cs
as + a'.g
h
Cs
at
h
Cs
a'.
h
"b
DETERMESTANTS
217
Develop the first determmant by minors with respect to the
first column, where we symbolize the minors of a^ + a'l, a^ + a'^,
ttg + a'l by Ai, A,, A^ respectively.
We have
by § 190,
aj + d's b, c.
= («! + «s'i)^i — (flSj + a'i)Ati +(ai8 + a',)At,
by the distributive law,
= UiAi — a^A^ + a^As + a'l^i — a'^A^ + a'^A,,
by § 190,
^1 ^1 Cl
0^2 ^2 ^2
^8 ^a ^8
+
Si
"i
s.
cs
6,
«8
The method of proof given is stppKcable to the case of any row
or column of a determinant of order n.
197. Vanishing of a determinant. For the solution of systems
of linear equations we shall make use of the
Theorem. If in the development of a determinant in terms
of the minors with respect to a certain column (or row) the ele-
ments of that column (or row) are replaced ly the elements of
another column (or row), the resulting development equals zero.
For example, we have
th
h
Cl
^1
Oj
h
c»
d.
a.
\
,"8
d.
ffli
h
Cl
d.
= a^Ai — Oj^j + tta^a — ttiAi,
(1)
where an A represents the minor of the a with the same sub-
script. We have to prove that if we replace the a's, for example,
by the ^s, the result equals zero, that is, that
biAi — b^Ai + b^A^ — biAi = 0.
(2)
218
ADVASrCED ALGEBRA
This expression (2) when written in determinant form evidently
would have the same form as the left-hand member of (1)' except-
ing that the first column would coifsist of Si , 5z , Jb > ** • We should
then have two identical columns of the determinant, which would
then equal zero (§ 194). Thus the development in (2) vanishes
identically. This method of proof is perfectly general.
CoKOLLAEY. The value of the determinant is unchanged if the
elements of any row (or column) are replaced by the elements of
that row (or column) increased by a multiple of the elements of
another row (or column).
Thus, for instance.
% &1 Ci
«i + nbi
h
<h
ttj ^2 Cj
=
eta + nb^
K
"2
0.8 ftg Cg
as + nbs
h
Cs
The proof follows directly from §§ 191, 196, and the preceding
theorem.
198. Evaluation by factoring. If in a determinant whose
elements are literal two rows or two columns become identical
on replacing a by b, then a — h is a factor of the development.
This appears immediately from § 160.
Illustration. Evaluate by factoring
1
b
b^
(1)
Since two columns become identical if a is replaced by b, a by c,
or b by c, then we have as a factor of the development
(a-b)(b-c)(c-a). (2)
To determine whether the signs in this product- are properly
chosen, that is, whether the development should contain a — 5 or
6 — a, we note that the term bo^ is positive in the development
of (1) and also positive in the expansion of (2). Evidently there
is no factor of (1) not included in (2).
DETERMINANTS
219
199. Practical directions for evaluating determinants. In
finding the value of a numerical determinant the object is to
reduce it to one in which as many as possible of the elements
of some row or column are zero. One should ask one's self the
following questions on attempting to evaluate a determinant :
First. Is any row (or column) equal to any other row (or
column) ? If so, apply § 191 for the case to = 0.
Second. Are the elements of any row (or column) multiples of
any other row (or column) ? If so, apply § 191.
Third. If we add (or subtract) a multiple of the elements of
one row (or column) to the elements of another, will an element
be zero ? If so, apply § 197.
EXERCISES
Evaluate the following determinants.
2
3
4
3
4
5
3
2
1
2
7
6
1
8
7
Solution : We observe that if we subtract each element of the first column
from the corresponding element of the second column, the new second column
has every element 1. A similar result is obtained by subtracting the last col-
umn from the third column. Thus, by § 191,
= 0.
2 3 4 3
2 113
4 5 3 2
4 112
12 7 6
1116
18 7
117
4
3
1
2
6
1
1
3
4
2
1
2
3
6
2
1
Solution : Multiplying the last column by 2 and the whole determinant by \
does not change the value of the determinant (§ 191). Thus
4 3 12
4 3 14
ff 1 1 3
_ 1
6 116
4 2 12
■"2
4 2 14
3 6 2 1
3 6 2 2
220
ADVANCED ALGEBRA
Subtracting the last column from the first column and developing, we
obtain
4
3
1
4
6
1
1
6
1
i
2
1
4
~ 2
3
6
2
2
3 14
116
2 14
16 2 2
3 14
116
2 14
Subtracting the last row from the first row,
5.
7.
11.
13.
15.
3 14
1 1 6
2 14
10
1 1 6
2 14
-m !i)=-^->--
a — d a 1
b-d b 1
c — d c 1
3 3 4 2
112 1
2 2 3 1
2 13 2
X
V
2
X
y
2
y
2
X
2
y
X
a
1
b
1
1
c
1
2
d
1
3
3
a
a
a
a
a
b
b
5
a
b
c
c
a
b
c
d
3
7
16
14
6
15
33
29
1
1
1
4
2
3
1
9
13
17
4
18
28
33
8
30
40
64
13
24
37
46
11
4.
10.
12.
14.
16.
a2 + 62
2ab
2db
o2 + &
4
6
8
8
1
1
2
1
2
3
4
1
2
1
3
4
110 1
10 11
111
1111
1110
110 1
10 11
111
2 111
12 11
112 1
1112
3 2 14
15 29 2 14
16 19 3 17
33 39 8 38
12
14
16
18
2
• 4
6
8
4
3
2
1
3
7
11
15
2a6 + &2 1
DETERMINANTS
221
200. Solution of linear equations. Suppose that we have given
n linear equations in n variables. We seek a solution of the equar
tions in terms of determinants. For simplicity, let ra = 4. Given
aix + biy + eT,s + diW=fi, (1)
a,x + b^y + Cja + d^w =fi, (2)
a^x + h^y + CjS + d^w =ft, (3)
a^x + biy + c^a + dtW =ft,. (4)
The coefficients of the variables taken in the order in which
they are written may be taken as forming a determinant D, which
we call the determinant of the system. Thus
«! Si Ci dy
^2 ^2 ^2 ^2 Tx
dj Ss Cj «fs
*1 ^4 "4 <^4
Symbolize by ^1, B^, etc., the minors of a-i, 63, etc., in this deter-
minant. Let us solve for x. Multiply (1), (2), (3), (4) by Ax, A^,
Ag, Ai respectively. We obtain
Aia^x + Aibiy + AiC^z + A^diW = A^fi,
A^a^x + A.J)iy + A^c^a + A^d^w = ^2/2,
A^a^x + A^b^y + ^8«8» + A^d^w = A^f^,
AiaiX + Aib^y + JI4C42: + j44<^4W = ^4/4.
If we add these equations, having changed the signs of the
second and fourth, the coefficient of x is the determinant D, while
the coefficients of y, s, w are zero (§ 197). The right-hand mem-
ber of the equation is the determinant D, excepting that the ele-
ments of the first column are replaced by /i, /z , /a , fi respectively.
Hence
« =
/l
h
Cl
d.
A
h
Cs
d.
A
h
Cb
d.
A
b.
"4
d.
ai
bi
Cl
d.
as
h
"2
d.
at
h
Ca
d.
a*
h
C4
d.
222
ADVANCED ALGEBRA
In a similar manner we can show that the value of any variable
which satisfies the equations is given by the following
Rule. The value of one of the variables in the solution of n
linear equations in n variables consists of a fraction whose
denominator is the determinant of the system and whose numer-
ator is the same determinant, excepting that the column which con-
tains the coefficients of the given variable is replaced by a column
consisting of the constant terms.
"When Z) = 0, we cannot solve the equations unless the numer-
ators in the expressions for the solution also vanish.
Illusteatiost. Solve for x
aa
, + 2by = l,
2by + 3cs = 2,
3os + 4:dw = 3,
4 dw + 5ax = i.
Eearranging, we obtain
ax + 2 by =1,
2
by + 3cz =2,
3cz + idw = 3,
5 ax
+ idw = A.
a 2b
a b
Z> =
2b
3c
3c 4d
= 24
b
c
c
d
5a
id
5a
d
1
b
1
b
24
2
b
c
1
c
1
c
3
c
d
3
c
d
-b
3
c d
4
d
4
d
4
d
a
b
a
b
b
c
24
b
c
b c
a
c d
c
d
c
d
-5b
d
5 a
d
-5b
d
DETEHMINANTS
223
1
c
-h
2
d
4
d
b
a
c
d
5e
d
be
d
ah
5c d
— 2 bed
— 4 abed
'2a'
201. Solution of homogeneous linear equations. Tlie equa^
tions considered in tlie previous section become homogeneous
(p. 116) if /i =/, =/3 =/, = 0. We have then
a^x + biy + CjS! + diW = 0,
a^x + b^y + C2« + d^w = 0,
'^s^ + bay + c^s 4- d^w = 0,
a^a; + b^y + CiZ + d^w = 0.
(I)
These equations have evidently the solution x = y = z=:w = 0.
This we call the zero solution. We seek the condition that the
coeifi-cients must fulfill in order that other solutions also may
exist. If we carry out the method of the previous section, we
observe that the determinant equals zero in the numerator of
every fraction which affords the value of one of the variables
(§ 191). Thus if D is not equal to zero, the only solution
of the above equations is the zero solution. This gives us
the following
Peinciple. a system of n linear homogeineous equations in n
variables has a solution distinct from, the zero solution only
when the determinant of the system, vanishes.
Whether a solution distinct from the zero solution always
exists when the determinant of the system equals zero we shall
not determine, as a complete discussion of the question would be
beyond the scope of this chapter.
Theorem. If x^, y^, z^, w^ is a solution of equations (I)
and k is any number, then hx^, hy^, hz^, kw^ is also a solution.
224
ADVANCED ALGEBRA
The proof of this theorem is evident on substituting kx^, etc.,
in equations (I) and observing that the number fc is a factor of
each equation. Thus if a system of n linear homogeneous equa-
tions has any solution distinct from the zero solution it has an
infinite number of solutions.
EXERCISES
Solve for all the variables :
x-\-y = a,
y + z = h,
1. z + tt = c,
u + v = d,
V + x = e.
z + y + w = a,
, z + w + x = b,
w + x + y = c,
x + y + z = d.
y + z + 5w = 11,
f. X-\- Z + 4:W = 11,
x + y + Siv = 11,
x + z-\-8y = S3.
x + y -i- z + w = 2i,
_ x+2y + Sz-9w = 0,
Sx — y — 6z + w = 0,
2x + ay-iz- 6w = 0.
4.
6.
a! + 3y = 19,
y + 3z = 8,
z + 3u = 7,
u + 3v = ll,
» + 3 a; = 15.
Sx + y + z = 20,
x + iy + 3w = S0,
6x + z + Sw = ^0,
82/ + 3z + 5m) = 50.
x — 2y + Sz — w = 6,
y -2z + 3w -x = 0,
z-2w+Sx-y = 0,
w — 2x + 3y — z = 5.
x + y + z + w^QO,
x + 2y + 3z-|-4m) = 100,
a; + 3y + 6« + 1010 = 150,
x + iy + 10z + 20 JO = 210.
CHAPTEE XIX
PARTIAL FRACTIONS
202. Introduction. For various purposes it is conTenient to
express a rational algebraic expression ^ V4 > § 11, as tlie sum of
several fractions called partial fractions, which have the several
factors of <^{x) as denominators and which have constants for
numerators. If we write <^ (x) = (a; — a) (a; — j8) ■ ■ • (a; — v), we
seek a means of determining constants A, B, ■■ •, N such that for
every value of x
<l>(x) x — a x — p 35— V ^-JL'-
If the degree of f(x) is equal to or greater than that of <f> (x),
we can write
,l>(x) <^(a!) X^-'
where Q(x) is the quotient and/i(a:) the remainder from dividing
f(x) hy 4> (x), and where the degree of /i (x) is less than that of
<f> (x). In what follows we shall assume that the degree of f(x)
is less than that of <^ (x). In problems where this is not the case
one should carry out the long division indicated by (2) and apply
the principle developed in this chapter to the expression corre-
sponding to the last term in (2).
203. Development when ^ («) = has no multiple rootsl Let
US consider the particular case "
f(x) ^ x + 1
<^(a;) (a; -1) (a; -2) (a; -3)'
We indicate the development required in form (1) of the last
section, . , „ „ - —
a; + l - ^ . -8 . C-
(a; - 1) (a; - 2) (x - 3) x - 1 a; - 2 a; - 3
22.5
226 ADVANCED ALGEBRA
where A, B, and C represent constants whieli we are to deter-
mine if possible. Tlie question arises immediately, Axe we at
liberty to make this assumption? Are we not assuming the
essence of what we wish to prove, i.e. the form of the expansion?
To this we may answer. We have written the expansion in form (1)
tentatively. We have not proved it and are not certain of its
validity. If, however, we are able to find numerical values of
A, B, and C which satisfy (1), we can then write down the actual
development of the fraction in the form of an identity.
If, on the other hand, we can show that no such numbers A,B,C
satisfying (1) exist, then the development is not possible. <,
Clear (1) of fractions,
X + 1 = A(x-2){x -Z) + B{x -l)(x -Z)+ C{x -l){x -2)
+IX°]:— {A + B + C)x'' — {6 A + 4: B + Z C)x +i& A + Z B + 2 c)'x
Since we seek values of A, B, and C for which (1) is identi-
cally, true for all values of x, equate coefficients of like powers of
a!,in the last equation (Corollary II, p. 174). We obtain
A+B+C = 0, (2)
-5A-iB-3C = l, (3)
(4)
(5)
(2)
6A+3B + 2C
= 1.
Add (4) to
(3)
and
we obtain
Adding we
obtain
A-B- C .
A +B+ C :
2A
A ;
= 2,
=
= 2
= 1
From (3) and (4) we obtain
B=-3, C = 2.
(SB - 1) (x - 2) (a: - 3) x-1 x-2 x-3
As a check we might clear of fractions and simplify. If equar
tibns (2),-(3), and (4) had been incompatible, we should have con-
cluded that we could not develop the fraction in form (1).
PARTIAL FRACTIONS 227
We assume now for the general case
<l>(x) = (x ~ a)(x — fi)--(x — V),
and that the n roots a, fi, ■■•, v are all distinct from each other.
Let lis consider the expression
^{x) X — a X — p X — V
where A, B, ■■■, N are constants. Let us assume for the moment
f(x)
the possibility of expressing j~r^ in terms of these partial frac-
tions. We shall now attempt to determine actual values A, B, ■■ ■,
N which satisfy such an identity. If we multiply hoth sides of ■
the identity by
<f>(x) = (x — a)(x — P)-,--(x — v), '
we obtain
f(x) = A(x-P)---(x — v) + B(x-a)---(x-v)-\
+ N(x-a)(x- p)-:
f(x) is of degree not greater than n — 1, and consequently when
written in the form of (1), p. 166, has not more than n terms. If
we multiply out' the right-hand member and collect powers of x,
we have an expression in x of degree n — 1. By Corollary II,
p. 174, this equation will be an identity if we can determine
values oi A, B, ■ ■ ■, N which make the coeflcients of x oji both
sides of the equation equal to each other. Hence we equate
coefficients of like powers of x and obtain n equations linear in
A,B, ■ ■ •, iV which we can treat as variables. These equations have
in general one and only one solution which we can easily deter-
mine. The values oi A, B,~- ■■, N obtained by solving these equa-
tions we can substitute for the numerators of the partial fractions
in (6). After making this substitution we can actually_clear of
fractions the right-hand member of (6) and check our "work by
'showing its identity with the left-hand member.
There is no general criterion that we have applied to (6) to
determine whether the n linear equations obtained by equating
coefSoient^' of x have any solution or not. Hence in this general
228 ADVANCED ALGEBRA
discussion it should be distinctly understood that assumption (6)
holds when and only when these equations are solvable. In
any particular ease we can find out immediately whether the
equations are solvable by attempting to solve them. If the num-
bers A, B, • • ; N do not exist, the fact will appear by our inability
to solve the linear equations. As a matter of fact, one and only
one solution always exists under the assimiption of this section.
If In (6) we assume that several of the symbols A, B, ■••, N stand for expres-
sions linear in z, as, for instance, ax + h,we should then have a larger number of
variables to determine than there are equations. Under these circumstances there
is an infinite number of solutions of the equations. Thus if we should seek to
express — — - as the sum of partial fractions where the numerators are not con-
<t>(x)
stants but functions of x, we could get any number of such developments.
We have the following
EuLE. Factor <f>(x) into linear factors, as
(x — a){x — /3)- ■•(x — v).
Write the expression
<f) (x) X — a X — /3 x — v
Multiply loth sides of the expression hy <f>(x), equate coefficients
of like powers of x, and solve the resulting linear equations for
A,B,--;N.
Beplace A, B, ■ • ■ , N hy these values and, check hy substituting
for X some numher distinct from a, P, • • • , v.
EXERCISES
Separate Into partial fractions :
CD«-2
1
(a!-l)(a;-2)a!
a? — 2 A B G
Solution : Assume : — = 1 u _, /n
(a;-l)(a-2)z z-\ x-% x ^'
Multiply by (a; - 1) (x - 2) x,
x^-'i=A(x- 2)x + B{x - Y)x + C(x - l)(a:- 2),
^-2 = (A+B-if C)x3 -(2A + B + 30X + 2C.
PARTIAL FRACTIONS 229
Equating coefficients of lilce
powers of x.
A + B + G = l,
24+B + 3C = 0,
2 C = - 2.
Hence
C = -l.
Solving
A + B = 2,
2A + B-Z,
we obtain
4 = 1
5 = l
x^ —
2
a; — 1 a; — 2 a;
(x-V)(x
-2)0!
Check: Let
a;:
= -1,
Substituting in (1),
-1
-6
--.2+-3 -l'
or
1
=-FI+'-l-'-l-
2 ^-^
3 x + l
■ a? + 3a + 2
" 2a:2-5a;-3
4. ^
3a;2-2a;-8
,. 6x
■ 6a;2-5a;-l
6. ^'^^
{??-i)(x-l)
y 2a;2-l
■ (a:i' + 3a: + 2)(a;-l)
g a:2-3a: + l
^)
9 x» + 4
■ (a:-l)(a;-2)(x-
(a;-2)(a; + 2)(x-l)
204. Development when <^(x) = has imaginary roots. In
the preceding section no mention has been made of any distinc-
tion between real and imaginary values of a, fi, ■■ ■, v. In fact
the jnethod given is valid whether they are real or imaginary. It
is, however, desirable to obtain a development in which only real
munbers appear.
Let US assume the development
^ = ^^ + -^ + - + ^^+— ' (1)
<^(a;) X — a x — p x — /j, x — v
where let us suppose that /n and v are the only pair of conjugate
imaginary roots of <^ (x), m and n being conjugate complex numbers.
230 ADVANCED ALGEBRA
Let ^/i, = a -[■ ib, V = a — ih.
Then adding the corresponding terms of (1), we obtain
'™ ra _ x(rn-\-n) — a{m + n') h(m — ri) .
x — a — ib x — a + ib (a; — a)^ + 6^ (a; — a)'' + b''
Since /x. and v are the only imaginary roots of <^ (a;) = 0, the last
term of (2) is real, as is also the entire right-hand member (§ 162).
Hence, letting the numerator
x(m + n) — a(m + ii) + ib (m — n)= Mx + N,
we have the development
</)(a;) x-a x-p {x - af + b'' ^'
By complete indnction we can establish this form of numerator
where there is any number of pairs of imaginary roots of <^(x) = 0.
We have proved the form (3) where there is one pair of imagi-
nary roots. Assuming the form where there are k pairs, we can
prove it similarly where there are h + 1 pairs. Hence we have the
Theorem. If (j> (x) is factorable into distinct linear and quad-
ratic factors, but the quadratic factors are not further reducible
f(x)
into real * factors, then ''/' ' is separable into partial fractions
of the form ^y /
A B , , Mx + N
1 TjH r ■
/8 0^ + fix + v
where a? + fix -{- v is an irreducible quadratic factor of ^ (x).
This theorem is of course true only under the condition that
the linear equations obtained in the process of determining the
constants are solvable. It turns out, however, that in this case
as in § 203 the linear equations obtained always have one and
only one solution provided that the roots are all distinct.
*A real factor is one whose coefficients are aU real numbers.
PARTIAL TRACTIONS 231
EXERCISES
Separate into partial fractions :
■ (x-l)(x' + x + l)'
c .• A »='' + 1 A ^ Bx+ G
Solution : Assume = +
(a; - 1) (a;2 + a; + 1) a; - 1 a;^ + s + 1
Multiply by {x - 1) (a;^ + a: + 1),
a;2 + 1 = 4 (a;2 + a; + 1) + (Ba; + 0) (a; - 1).
Collecting like powers of x,
3? + 1= {A + B)x^ + (A - B + C)x + A - C.
Equating coefficients of x,
A + B = l, (1)
A-B+G = 0, (2)
A-C = l. (3)
Add (2) and (3) and solve with (1),
2A-B=1
A + B = l
34 = 2,
or 4 = 1-
Substituting in (1), f + B = 1,
or B = i.
Substituting in (3), | - G = 1,
or C = -J.
Thus ^' + ' _ 2 , .-1
(a;-l)(a;2 + a; + l) 3(a;-l) 8a;2+3a; + 3
Check: Let a; = — 1.
Substituting,
2 2 -2
-2-1 3-2 3-3 + 3
2 2
Reducing, -1 = ---^=-!.
2 x^ + x + 1 3 a;2+l
a;» + 4x
a;2 + 4
a;8_2a;2 + 3a;-
a;
2
(a; + 3) (2x2 -a;
x2 + l
-4)
4. A"": — r- 5.
7.
X* + x2 + 1
5x8-1
x* + 6x2 + 8'
1
x3 + 3x2-2x-16
HiHT. Factor by synthetic divi-
(x-l)(3x2 + x + 6) ' Bion (see p. 178) .
232 ADVANCED ALGEBRA
205. Development when ^(aj) = (ae — o)". In this case the
method given in the previous sections fails, as the equations for
determining the values of the numerators are incompatible. If
<j,(x) («-«)» ' ^ ^
we can separate into partial fractions as f oUows.
Let X — a = y, that is, cc = y + a, and substitute in (1). We
obtain after collecting powers of y
2/" y y^ ' y" '
■where the A's are constants. Replacing y by a; — «, we have the
following development :
f(x) _ ^0 ^1 ^3 , ^
(a; - «)» a; - a "*" (a; - a)" "'' (a; - «)» "^ ■ ■ ■ "^ (a; - «)»
EXERCISES
Separate into partial fractions :
- ic' + Za-l
(x-3)a
solution: Assume __^ = —^ + _^^ -, ^-_^. (1)
Multiply by (x - 3)',
a:" + 2a - 1 = ^(cc^ - 6a; + 9) + B(x - 3) + C.
Collecting powers of x,
= Ai)fi + {B-6A)x + 9A-SB+C.
Equating coefficients of x,
A = l,
B-6A = 2,
9A-SB + C = -1.
Solving, B = 8, C = U.
a^ + 2a;-l 1 , 8 ,14
Hence -— — — = + — +
(x-3)8 x-3 {x-3)2 (x-3)s
Check : Let x = 1.
o V .-^ .- ■ /,x 2 1 , 8 14 1 „ 14
Substitutmg m (1), -— = -_ + --_=__ + 2 -—,
1 1
PARTIAL FRACTIONS 233
(a -2)8" ■ (2 a + 1)2' ■ (a -4)8'
d' + x + l e x-a „ 2s2 + 3a + l
(2a! -1)4 (ax + hy (3a; -2)8
206. General cage. Wten </> (a;) = has real, complex, and
multiple roots, we may use all the previous methods simultane-
ously. Hence for this case we assume the expansion
Zfel
(» — «)••• (Axc'-H jua; + v) • • ■ (x — t)
,*
^ + ... + ^^£±i^4-... + ^^ + -+-
\X^-\- flM + V X — T (X — t)'
,k
EXERCISES
Separate into partial functions :
, !C< + 2a' + 18a-18
■ (a;-l)(a;2+x + l)(a-2)2'
Solution :
a* + 212 + 18a; -18 ABx + CD , E
{x-l)(x^ + x + l)(x-2Y x-1 x^ + x + 1 x-2 (a-2)2
= A(i? + x + l)(x-2Y + (Bx + C!)(x-2f(x-l)
+ D{x-l)(x-2)(x' + x + l)
+ E(x -l){x^ + X + 1)
= {A + B+I))x*+(-SA-5B + C-2D+E)x«
+ (A + S B ^ 5C)ai' + (- iB + 8 G - D)x
+ (iA-iG + 2I)-E).
Equating coefficients of lilte powers of x,
A+ B + D =1, (1)
-3A-5B+ G-2D + E = 0, (2)
A + 8B-6C =2, (3)
-4B + 8C- D =18, (4)
iA -iG + 2D-I! = -lS. (5)
Adding (2) and (5), (1) and (4), we have, together with (8),
A-&B-3C = -1S, (6)
.il-3B + 8C = 19, (7)
A + 8B-6C = 2. (8)
234 ADVANCED ALGEBRA
Adding all three equations, we find
3 A = 3, or 4 = 1.
Substituting in (3) and (7) and solving, we find C = 3,B = 2. Substituting
in (1), we find D = - 2. Similarly, from (5), £ = 6.
Thus ^ + 2»=" + 18=.-
-18
a; -2)2
1
x-l
2a: + 3 2 6
x'^ + x + X x-2 (x-2)2
Check: Leta; = — 1.
Substituting, — — -
— 2 •
33
1-9
^-\
1 2 6
I -3 + 9'
or
^ =
^-\■
2 ^' + ^
3
4a!2
"■ a;(a;-l)8
1-!k4
. a;8 + 2a;2 + l
5
5a; + 12
x(x-l)^
a;(a;2 ^. 4)
g 2a;' + a: + 3
7
43X-11
(x" + If
30(a;2 - 1)
' 8. ^-^ .
(x + lY(x + 2)
9
a;3 - a; - 1
a;* -16
10 "^ + ^
{x^-l){x + 2)
<
11.
a;2 + 6a;-8
a;' -4 a;
^2 x-^-Qx + l
13
2
{x-8)(a;-9)
(a;2 + a; + 3)(2a; + l)
j^. 2a;2-3a;-3
15
a;s + 2a;2-3a; + l
■ (s-l)(a;2-2a;+5)
(x + 3)(a;2-4s+6)
j^g 30a; -17
_.
17.
13 - 5a;
2x-3)(6a;2-5a; + l) (a;^- 3a; + 2)(s - 3)
CHAPTEE XX
LOGARITHMS
207. Generalized powers. If h and e are integers, we can easily
compute ¥. When c is not an integer but a fraction we can com-
pute the value of b" to any desired degree of accuracy. Thus if
5 = 2, c = 1^, we have 2* = V2' = Vs, which we can find to any
number of decimal places. If, however, the exponent is an irra^
tional number as V2, we have shown no method of computing
the expression. Since, however, V2 was seen (p. 55) to be the
limit approached by the sequence of numbers
1, 1.4, 1.41, 1.414, ■ • •-,
it turns out that 5^^ is the limit approached by the numbers
5'-, 5'-^»6"S 5"», ••-.
The computation of such a number as 5^-^' would be somewhat
lii lOT/
laborious, but could be performed, since 5'*' = 6"* = v6"'. Thus
it is a root of the equation a;""' = 5"^, and could be found by
Horner's method, p. 197.
We see in this particular case that 5^^ is the limit approached
by a sequence of numbers where the exponents are the successive
approximations to V2 obtained by the process of extracting the
sqflare root. In a similar mariner we could express the meaning
of b°, where 6 is a positive integer and o is any irrational number.
Assumption. We assume that the laws of operation which we
have adopted for rational exponents hold when the exponents are
irrational.
Thus &=.6d=6c + d^ — — ^c-d^ {bd)c—(J,c)d—liCd^
where c and d are any numbers, rational or irrational.
235
236 ADVANCED ALGEBRA
208. Logarithms. We have just seen that when b and e are
given a number a exists such that 6° = a. We now consider the
case where a and h are given and c remains to be found. Let
a = 8, & = 2. Then if 2° = 8, we see immediately that e = 3 satis-
fies this equation.. If a = 16, 6 = 2, then 2' = 16 and c = 4 is the
solution. If a = 10, 6 = 2, consider the equation 2° = 10. If we
let c = 3, we see that 2' = 8. If we let c equal the next larger
integer, 4, we see 2* = 16. If then any number c exists such that
2" = 10, it must evidently lie between 3 and 4. To prove the exist-
ence of such a number is beyond the scope of this chapter, but
we make the following
Assumption. There always exists a real numher x which
satisfies the equation
b' = a, (1)
where a anid i are positive numbers, provided b ^ 1.
Since any real number is expressible approximately in terms of
a decimal fraction, this number a; is so expressible.
The power to which a given nurkher called the base must be
raised to equal a second number, is called the logarithm of the
second number.
■ In (1) X is the logarithm of a for the base b.
This is abbreviated into
X = logjtt. (2)
Thus since
28 = 8, 102=100, 3-2=J, 4» = 1,
we have
3 = log2 8, 2 = logio 100, - 2 = lo^ i, = logi 1.
The number a in (1) and (2) is called the antilogarithm.
EXERCISES
1. In the following name the base, the logarithm, and the antilogarithm,
and write in *^m (2).
(a) 36 = 729.
Solution : 3 = hase, 6 = logarithm, 729 = antilogarithm, logs 729 = 6.
(b) 2* = 16. ~~ " ~~~~-N ",-g) 38 = 27.
LOGARITHMS 237'
2. Find the logarithms of the following numbers for the base 3 :
81, 243, 1, I, tV.
3. For base 2 find logarithms of 8, 128, J, ^.
4. What must the base be when the following equations are true ?
(a) log49 = 2. (b) log81 = 4.
(c) log 225 = 2 . (d) log 625 = 4.
209. Operations on logarithms. By means of the lav expressed
in the Assumption, § 207, we arrive at principles that have made
the use of logarithms the most helpful aid in computations that
is known.
Theorem I. The logarithm of the product of two numbers is
the sum of their logttrithms.
Let logjft = X,
logjc = y.
Then hy (1) and (2), p. 236, If = a,
by = c.
Multiply (by Assumption, § 207),
b''+y = a-e,
or by (1) and (2), logj a. • c = x + y.
Theorem II. The logarithm of the nth power of a number is
n- times the logarithm, of the number.
Let log;, a — X,
or 6* = as.
Eaise both sides to the reth p6wer,
(Jfy = b'^ = a",
or logj a" = nx.
Example. Iogiol0p = 2,
logiolOOO = 3T^
By Theorem I, logio 100,000 = 5,
which is evidently true, since 10^ = 100,000.
238 ADVANCED ALGEBRA
Theorem IIL The logarithm of the quotient of two numbers
is the difference between the logarithms of the numbers.
Let logja = X,
logjO = y.
Tlien h" = a,
W = c.
Dividing, y-y = -,
G
a
or logj - = x — y:
Theorem IV. The logarithm of the real nth root of a nur.i-
ber is the logarithm of the number divided by n.
Let logja = a:,
or y = a.
L X
Extract tlie wth root, Qfy. = J» = -yfa,
log, 7^=^.
EXERCISES
Givenlogio2 = .301, logio5 = .699, logioT = .8451, find
1. log(\/7S- VS).*
Solution :
By Theorem I, log (VT^ • Vs) = log -v^ + log V5.
By Theorems III and IV, = | log 7 + i log 5.
Now flog? = f(. 8451) = .50706,
and \ log 5 = ^ (. 699) = .3495
Adding, I log 7 + i log 5 = log ( v^ . VB) = -.86656
2. log 40. 3. log 28.
Hint. Let 40 = 8-5= 23-5.
4. log 140. 5. logv^lO.
6. log V^. 7. log ( V8 . 4^ ■ ^.
8. log ( V5 ■ 78). 9. log (-t^ . Vii ■ V^.
* Where no base Is written it is assumed that the base. JO^s employed.
LOGARITHMS 239
210. Common system of logarithms. For purposes of compu-
tation 10 is taken as a base, and unless some other base is indi-
cated we shall assume that such is the case for the rest of this
chapter. We may write as follows the equations which show the
numbers of which integers are the logarithms.
Since 10« = 100,000 we have log 100,000 = 6.
10^ = 10,000
log 10,000 =4.
10» = 1000
log 1000 = 3.
10' = 100
log 100 = 2.
10> = 10
log 10 'i =1.
10» = 1
log 1 =0.
10- = .1
log.l =-1.
10- = = .01
log .01 =-2.
10-' = .001
log .001 =-3.
etc.
etc.
Assuming that as x becomes greater log x also becomes greater,
we see that a number, for example, between 10 and 100 has a
logarithm between 1 and 2. In fact the logarithm of any number
not an exact power of 10 consists of a whole-number part and a
decimal part.
Thus since lOS < 342K 10*,
log 3421 = 3. -I- a decimal.
Since 10- 3 <. 0023 < 10-2,
log .0023 = - 3. -I- a decimal.
The whole-number part of the logarithm of a number is called
the characteristic of the logarithm.
The decimal part of the logarithm . of a number is called the
mantissa of the logarithm.
The characteristic of the logarithm of any number may be seen
from the above table, from which the following rules are imme-
diately deduced.
■ The characteristic of the logarithm, of a number greater^ than unity
is one less than the number of digits to the left of its decimal point.
Thus the characteristic of the logarithm of 471 is 2, since 471 is between 100
and.lOOO; of 27.93 is 1, since this number is.between 10 and 100; of 8964.2 is 3,
since this number is between 1000 and 10,000.
240 ADVANCED ALGEBRA
The characteristic of the logarithm of a number less than 1
is one greater negatively than the numher of zeros preceding the
first significant figure.
Thus the characteristic of the logarithm o£ .04 is — 2; of .006791 is —3; of
.4791 is - 1.
It must constantly be kept in mind that the logarithm of a
number less than 1 consists of a negative integer as a character-
istic plus a positive mantissa. To avoid complication it is desir-
able always to add 10 to and subtract 10 from a logarithm when
the characteristic is negative. Thus, for instance, instead of writ-
ing the logarithm - 3 -f .4672 we write 10 — 3-1- -4672 — 10, or
7.4672 — 10. This is convenient when for example we wish to
divide a logarithm by 2, as by Theorem IV, § 209, we shall wish
to do when we extract a square root. Since in the logarithm
— 3 + .4672 the mantissa is positive, it would not be correct to
divide — 3.4672 by 2, as we should confuse the positive and
negative parts. This confusion is avoided if we use the form
7.4672 - 10, and the result of division by 2 is 3.7336 - 5, or
8.7336 — 10. The actual logarithm which is the result of this
division is — 2 + .7336.
Theorem. Numbers with the same significant figures which
differ only in the position of their decimal points have the sarne
mantissa.
Consider for example the numbers 24.31 and 2431.
Let lO"' .= 24.31.
Then x = log 24.31.
If we multiply both numbers of this equation by 100, we have
10n(y = 10^+2 = 2431,
or x + 2 = log 2431.
Thus the logarithm of one number differs from that of the other
merely in the characteristic. In general numbers with the same sig-
nificant figures are identical except for multiples of 10. Hence their ,
logarithms differ only by integers, leaving their mantissas the same.
Thus if log47120. = 4.6732, log47.12 = 1.6732, and log .004712 = - 3.6732, or
7.6732 - 10,
LOGARITHMS 241
EXERCISES
If log2 = .3010, logs = .4771, log7 = .8451, find
1. logVeoo.
Solution : log V600 = log V20 • 30 = J log 20 + J log 30.
By the preceding theorem, log 20 = 1.3010, log30 = 1.4771.
i log 20 =
.6505
* log 30 =
.73855
By § 209,
log -^600 =
1.38905
2. log .06.
3.
log(210)».
4. log 5.4.
5. log (4. 2)*.
6.
, 9.6
'■ '"'S^'
o , 567
, 13.23
«■ >°^I6
, 9.
^°^1.28-
211. Use of tables. A table of logaritlims contains the man-
tissas of the logarithms of all numbers of a certain number of
significant figures. The table found later in this chapter gives
immediately the mantissas for all numbers of three significant
figures. In the next section a method is given for finding the
mantissa for a number of four figures. Hence the table is called
a four-place table. Before every mantissa in the table a decimal
point is assumed to stand, but in order to save space it is not
written. To find the logarithm of a number of three or fewer
significant figures we apply the following
EuLE. Determine the characteristic hy rules in § 210.
Find in column N the first two significant figures of the num-
ber. The mantissa required is in the row with these figures.
Find at the top of the page the last figure of the number.
The mantissa required is in the column with this figure.
When the first significant figure is 1 we may find the loga-
rithm of any number of four figures by this rule from the table
on pp. 248, 249 if we find the first three instead of the first two
figures in column N. ^
Thus the log516. = 2.7126, log .00281 = - 3.4487, log 7400. = 3.8692,
logapo. = 2.7782, logSO. = 1.6990, logl.OO = .6021.
242 ADVANCED ALGEBRA
EXERCISES
Pind tlie logaritlims of the following :
1. 3.
2. 303.
3. .024.
4. 347.
5. .0333.
6. 1.011.
7. .202.
8. .0029.
9. .0001.
10. .00299.
11. 68400.
12. .0201.
212. Interpolation. We find by the preceding rule that
log 2440 = 3.3874, while log 2450 = 3.3892. If we seek the loga-
rithm of a rnimber between 2440 and 2450, say that of 2445,
it would clearly be between 3.3874 and 3.3892. Since 2445 is just
halfway between 2440 and 2450, we assume that its logarithm is
halfway between the two logarithms. To find log 2445, then, we
look up log 2440 and log 2450, take half (or .6) their difference,
and add this to the log 2440. This gives
log 2445 = 3.3874 + .5 x .0018 = 3.3883.
If we had to find log 2442 we should take not half the difference but
two tenths of the difference between the logarithms of 2440 and
2460, since 2442 is not halfway between them but two tenths of the
way. This method is perfectly general, and we may always find the
logarithm of a number of more than three figures by the following
EuLE. Annex to the ■pro'per characteristic the mantissa of
the first three significant figures.
Multiply the difference between this mantissa and the next
larger mantissa in the table (called the tabular difference and
denoted by D) by the remaining figures of the number preceded
by a decimal point.
Add this product to the extreme right of the logarithm of the
first three figures,, rejecting all decimal ploAies beyond' the fourth.
In this process of interpolation we have assumed and used the principle that
the increase of the logarithm is proportional to the increase of the number. This
principle is not strictly true, though for numbers whose first significant figure
is greater than 1 the error is so small as not to appear in the fourth place of the
logarithm. For numbers whose first significant figure is less than 2 this error
■ would often appear if we found the fourth place by interpolation. For this reason
the table on pp. 248, 249 gives the logarithms of all such numbers exact to four
figures, and in this part of the table we do not need to interpolate at all
LOGARITHMS
243
EXERCISES
Find the logarithms of the following :
1. 63.924.
Solution :
, „„ „ , „ -, Tabular difference = 7
log63.9= 1.8055 24
2 -^
log63.924 = 1.8057 "_
We add 2 to 1.8055 rather than 1, because 1.68 ia nearer 2 than 1. In general
B take the nearest integer.
2. 269.4.
3. 1001. 4. 62230.
5. 392.8.
6. 9.365. 7. 20060.
8. .4283.
9. .3101. 10. 9.999.
11. 82.93.
12. .05273. 13. 5.7828.
14. .003011.
15. .002156. 16. 3.1416.
17. 275.4 X 1.463.
Solution :
I0g275.4 = 2.4399 log275=2.4393 2)= 16
Iogl.463 = 0.1652 6 .4
Bv Theorem II. S a
09. loff ^275.4 X 1.4B.<?'l = 2.fi0fiT 2.4399 6.4
18. 374.3 X 1396. 19. .1.46 x 237.2.
20. 469.1 X 63.92. 21. 47320. x
22. :2!!?1.
38.46
Solution : log . 03724 = 8. 5710 - 10
8.5710-
log 38.46 = 1.5850
6.9860 -
10
23.
3.467
.2364'
24.
■06792
5.128 '
log .0372 =8.5705 -10 Z)= 12
5 _A
8.5710-10 4.8
log 38.46= 1.5843 Z>= 12
7 _^
1.6850 7.2
213. Antilogarithms. We can now find the product or quotient
of two numbers if we are able to find the number that corresponds
to a given logarithm.
For this process we have the following
liULE. If the mantissa is found exactly in the table, the first
two figures of the corresponding number are found in the column
N of the same row, while the third figure of the number is found
at the top of the column in which the mantissa is found.
Place the decimal point so that the rules in § 210 are fulfilled.
244 ADVANCED ALGEBRA
EXERCISES
Find the antilogarithms of the following:
1. 3.7419.
Solution: We find the mantissa .7419 in the row which has 55 in col-
umn N. The column in which .7419 is found has 2 at the top. Thus the
significant figures of the antilogarithm are 552. Since the characteristic is 3,
we must by the rule in § 210 have four figures to the left of the point.
Thus the number sought is 5520.
2. 1.3874. 3. 2.7050. 4. .6785.
5. 2.8414.* 6. 5.8831. 7. 1.5752.
8.9.9112-10. 9.3.7251. 10.5.3997.
If the mantissa of the given logarithm is between two man-
tissas in the table, we may find the antilogarithm by the following
Rule. Write the number of three figures corresponding to the
lesser of the two mantissas between which is the given mantissa.
Subtract this mantissa _ from the given mantissa, and divide
this number by the tabular difference to one decimal place.
Annex this figure to the three already found, and place the
decimal point as the rules in § 210 require.
It shonld be kept in mind that we may always add and subtract
any integer to a logarithm. This is useful in two cases :
First. When we wish to subtract a larger logarithm from a
smaller ;
Second. When we wish to divide a logarithm by an integer
that is not exactly contained in the characteristic.
Both these processes are illustrated in exercise 2 Q) following.
EXERCISES
1. Find the antilogarithms of the following :
(a) 2.3469.
Solution : The mantissa 3469 is between 3464 and 3483. Hence D = 19.
The mantissa 3464 corresponds to 222. To find the fourth significant figure
of the antilogarithm, divide 3469 - 3464 = 5 by D = 19. Since 5 -=- 19 = .26,
we annex 3 to 222. Hence the antilogarithm = 222.3.
* We write - 2 + .8414 in the form 2.8414 to save space and at tHe same time to recall
tlie fact tliat the mantissa is positive. *
LOGARITHMS
245
(b) 4.3147. (o) 1.5271. (d) 1.4216.
(e) 1.6423. (f) 2.8791. - (g) .7214.
2. Perforin the following operations by logarithms.
1375 X .06423
(a)
76420
Solution :
log 1375= 3.1383
log .06423 = 8.8077 - 10
11.9460 - 10
log 76420 = 4.8832
Subtracting (Theorem III, § 209), log result = 7.0628 - 10
result = .001156.
Adding (Theorem I, § 209),
(b) (W-
(h) 8308 X .0003769.
8.371 X 834.6
(d) 5871 ^ i
(f) (3f J)*-". (g) 7065 -=- 5401.
(i) 3410 X .008763.
37.42 X 11.21
7309
38.47
87 X-
(3) ;:j • W
\ 98080000
Solution: log 87 = 1.9395
By Theorem rv, § 209, J log 7 194 = 1.9285
, Adding, ' =13.8680-10
log 98080000 = 7.9916
By Theorem III, § 209, 3 )25.8764 - 30
log result = 8.6255 - 10
result = .04222.
Since in the subtraction in this problem we have to subtract 7 from 3,
we add and subtract 10 to the minuend to avoid a negative logarithm. Since
in the division by 3 we would have a remainder in dividing — 10 by 3, we
add and subtract 20 so that 3 may be exactly contained in 30, the negative
part of the logarithm.
(m) V7.
(n) V:06.
(P) '^i.
(q) (.21)5.
(s) </:o3.
(t) vlOO.
(v) V.00614.
(w) V?H-
(y) ^•■
Viii-
(o) V(.043)8.
(r) m -^.
(u) V(1.563)«.
(X) 7o. 9 VH-
(z) 7-4T VS.
214. Cologarithms. The logarithm of the reciprocal of a num-
ber is called its cologarithm. When a computation is to be made
246
ADVANCED ALGEBRA
in wMch several numbers occur in the denominator of a fraction,
the subtraction of logarithms is conveniently avoided by the use
of cologarithms. By our definition we have
colog 25 = log 5>j = log 1 - log 25, Theorem III, § 209
log 1 = 10. - 10
log 25 = 1.3979
colog 25= 8.6021-10
Thus in dividing a number by 25 we may subtract the logarithm
of 25, or what amounts to the same thing, add the logarithm of
5^, which is by definition the cologarithm of 26.
EuLE. The cologarithm, of any number is found iy subtract-
ing its logarithm from 10 — 10.
In the process of division subtracting the logarithm of a num-
ber and adding its cologarithm are equivalent operations.
EXERCISES
Compute, using cologarithms.
8 X 62.73 X .052
1.
X 8.793
Solution :
log8= .9031
log 62. 73= 1.7975
log. 052= 8.7160-10
colog 56= 8.2518-10
colog 8.793 = 9.0559-10
log result = 27.7243 - 30
result = .005299
9.
109 , /76
V38.462 - 16.382.
5086 (.0008769)8
9802 (.001984)* '
4:
4
693 X .04692
03841 X (569.8)2'
(.068)8 X 421.6 X 8°
VeOx .045.(200.1)^"
3. V1572 - 872.
Hint. 1572-872 = (157 +87) (157 -87)
= (244) (70).
6. V(27.5)2 - (3.483)2.
96)^ X 86.51
09263 -v^
10.
/(5a
8 /3.1416 X (5.638)2
■ \ (75)*
LOGARITHMS 247
215. Change of base. We have seen that the logarithm of a
number for the base 10 may be found to four decimal places in
our tables. It is occasionally necessary to find the logarithm of a
number for a base different from 10. For the sake of generality,
we assume that the logarithms of all numbers for a base h are
computed. We seek a means of finding the logarithm of any
number, as x, for the base o ; that is, we seek to express log^x in
terms of logarithms for the base h.
Suppose logo a; = a, that is, <f = x.
Take the logarithm of this equation for the base h, and we have
logjc^ = » log^c = logjs;.
Then
^ log,a;
logjc
If we let
M = log^c,
we have
logjO!
^~ M
(1)
This number M does not depend on the particular number x,
but only on the two bases. From (1) we see that we can find the
logarithm of any number for the base c by dividing its logarithm
for the base b by M. The number M is called the modulus of the
new system with respect to the original one. •
EuLE. To find the logarithm of a nurrbber for a new hase c,
divide the common logarithm ly the modulus of the system whose
hase is c.
EXERCISES
Find:
1. logs 21.
Solution: log,21->''S-^;- Iff -2.771.
logioS .4771
2. log6 6. 3. log2l5.
4. l0gl6-2.
5. logs 167. 6. logi8 237.
7. 10g2.i6l.41.
248
ADVANCED ALGEBRA
N.
1
2
3
4
5
6
7
8
9
100
0000
0004
0009
0013
0017
0022
0026
0030
0035
0039
101
102
103
104
105
106
107
108
109
0043
0086
0128
0170
0212
0253
0294
0334
0374
0048
0090
0133
0175
0216
0257
0298
0338
0378
0052
0095
0137
0179
0220
0261
0302
0342
0382
0056
0099
0141
0183
0224
0265
0306
0346
0386
0060
0103
0145
0187
0228
0269
0310
0350
0390
0065
0107
0149
0191
0233
0273
0314
0354
0394
0069
0111
0154
0195
0237
0278
0318
0358
0398
0073
0116
0158
0199
0241
0282
0322
0362
0402
0077
0120
0162
0204
0245
0286
0326
0366
0406
0082
0124
0166
0208
0249
0290
0330
0370
0410
110
0414
0418
0422
0426
0430
0434
0438
0441
0445
0449
111
112
113
114
115
116
117
118
119
0453
0492
0531
0569
0607
0645
0682
0719
0755
0457
0496
0535
0573
0611
0648
0686
0722
0759
0461
0500
0538
0577
0615
0652
0689
0726
0763
0465
0504
0542
0580
0618
0656
0693
0730
0766
0469
0508
0546
0584
0622
0660
0697
0734
0770
0473
0512
0550
0588
0626
0663
0700
0737
0774
. 0477
0515
0554
0592
0630
0667
0704
0741
0777
0481
0519
0558
0596
0633
0671
0708
0745
0781
0484
0523
0561
0599
0637
0674
0711
0T48
0785
0488
0527
0565
0603
0641
0678
0715
0752
0788
120.
0792
0795
0799
0803
0806
0810
0813
0817
0821
0824
121
122
123
124
125
126
127
128
129
0828
0864
0899
0934
0969
1004
1038
1072
1106
0831
0867
0903
0938
0973
1007
1041
1075
1109
0835
0871
0906
0941
0976
1011
1045
1079
1113
0839
0874
0910
0945
0980
1014
1048
1082
1116
0842
0878
0913
0948
0983
1017
1052
1086
1119
0846
0881
0917
0952
0986
1021
1055
1089
1123
0849
0885
0920
0955
0990
1024
1059
1093
1126
0853
0888
0924
0959
0993
1028
1062
1096
1129
0856
0892
0927
0962
0997
1031
1065
1099
1133
0860
0896
0931
0966
1000
1035
1069
1103
1136
130
1139
1143
1146
1149
1153
1156
1159
1163
1166
1169
131
132
133.
134
135
136
137
138
139
1173
1206
1239
1271
1303
1335
1367
1176
1209
1242
1274
1307
1339
1370
1402
1433
1179
1212
1245
1278
1310
1342
1374
1405
1436
1183
1216
1248
1281
1313
1345
1377
1408
1440
1186
1219
1252
1284
1316
1348
1380
1411
1443
1189
1222
1255
1287
1319
1351
1383
1414
1446
1193
1225
1258
1290
1323
1355
1386
1418
1449
1196
1229
1261
1294
1326
1358
1389
1421
1452
1199
1232
1265
1297
1329
1361
1392
1424
1455
1202
1235
1268
1300
1332
1364
1396
1427
1458
140
1461
1464
1467
1471
1474
1477
1480
1483
1486
1489
141
142
143
144
145
146
147
148
149
1492
1523
1553
1584
1614
1644
1673
1703
1732
1495
1526
1556
1587
1617
1647
1676
1706
1735
1498
1529
1559
1590
1620
1649
1679
1708
1738
1501
1532
1562
1593
1623
1652
1682
1711
1741
1504
1535
1565
1596
1626
1655
1685 ,
1714
1744
1508
1538
1569
1599
1629
1658
1688
1717
1746
1511
1541
1572
1602
1632
1661
1691
1720
1749
1514
1544
1575
1605
1635
1664
1694
1723
1752
1517
1547
1578
1608
1638
1667
1697
1726
1755
1520
1550
1581
1611
1641
1670
1700
1729
1758
150
1761
1764
1767
1770
1772
1775
1778
1781
1784
1787
N.
1
2
3
4
5
6
7
8
9
LOGARITHMS
249
N.
1
2
3
4
5
6
7
8
9
ISO
1761
1764
1767
1770
1772
1775
1778
1781
1784
1787
151
152
133
154
155
156
157
158
-159
1790
1818
1847
1875
1903
1931
1959
1987
2014
1793
1821
1850
1878
1906
1934
1962
1989
2017
1796
1824
1853
1881
1909
1937
1965
1992
2019
1798
1827
1855
1884
1912
1940
1967
1995
2022
1801
1830
1858
1886
1915
1942
1970
1998
2025
1804
1833
1861
1889
1917
1945
1973
2000
2028
1807
1836
1864
1892
1920
1948
1976
2003
2030
1810
1838
1867
1895
1923
1951
1978
2006
2033
1813
1841
1870
1898
1926
1963
1981
2009
2036
1816
1844
1872
1901
1928
1966
1984
2011
2038
160
2041
2044
2047
2049
2052
2055
2057
2060
2063
2066
161
162
163
164
165
166
167
168
169
2068
2095
2122
2148
2175
2201
2227
2253
2279
2071
2098
2125
2151
2177
2204
2230
2256
2281
2074
2101
2127
2154
2180
2206
2232
2258
2284
2076
2103
2130
2156
2183
2209
2235
2261
2287
2079
2106
2133
2159
2185
2212
2238
2263
2289
2082
2109
2136
2162
2188
2214
2240
2266
2292
2084
2111
2138
2164
2191
2217
2243
2269
2294
2087
2114
2140
2167
2193
2219
2245
2271
2297
2090
2117
2143
2170
2196
2222
2248
2274
2299
2092
2119
2146
2172
2198
222.6
2251
2276
2302
170
2304
2307
2310
2312
2315
2317
2320
2322
2325
2327
171
172
173
174
175
176
177
178
179
2330
2355
2380
2405
2430
2455
2480
2504
2529
2333
2358
2383
2408
2433
2458
2482
2507
25S1
2335
2360
2385
2410
2435
2460
2485
2509
2633
2338
2363
2388
2413
2438
2463
2487
2512
2536
2340
2365
2390
2415
2440
2466
2490
2614
2638
2343
2368
2393
2418
2443
2467
2492
2616
2541
2345
2370
2395
2420
2445
2470
2494
2519
2543
2348
2373
2398
2423
2448
2472
2497
2621
2645
2350
2375
2400
2425
2450
2476
2499
2524
2548
2353
2378
2403
2428
2463
2477
2502
2626
2550
180
2553
2555
2558
2660
2662
2565
2567
2570
2572
2574
181'
182
183
184
185
186
187
188
189
2577
2601
2625
2648
2672
2695
2718
2742
2765
2579
2603
2627
2651
2674
2697
2721
2744
2767
2582
2605
2629
2653
2676
2700
2723
2746
2769
2584
2608
2632
2666
2679
2702
2725
2749
2772
2586
2610
2634
2658
2681
2704
2728
2751
2774
2689
2613
2636
2660
2683
2707
2730
2753
2776
2691
2615
2639
2662
2686
2709
2732
5765
2778
2694
2617
2641
2665
2688
2711
2736
2768
2781
2596
2620
2643
2667
2690
2714
2737
2760
2783
2598
2622
2646
2669
2693
2716
2739
2762
2786
190
2788
2790
2792
2794
2797
2799
2801
2804
2806
2808
191
192
193
194
195
196
197
198
199
2810
2833
2856
2878
2900
2923
2945
2967
2989
2813
2835
2858
2880
2903
2925
2947
2969
2991
2815
2838
2860
2882
2905
2927
2949
2971
2993
2817
2840
2862
2885
2907
2929
2951
2973
2995
2819
2842
2865
2887
2909
2931
2953
2975
2997
2822
2844
2867
2889
2911
2934
2956
2978
2999
2824
2847
2869
2891
2914
2936
2958
2980
3002
2826
2849
2871
2894
2916
2938
2960
2982
3004
2828
2851
2874
2896
2918
2940
2962
2984
3006
2831
2853
2876
2898
2920
2942
2964
2986
3008
200
3010
3012
3015
3017
3019
3021
3023
3026
3028
3030
N.
1
2
3
4
5
6
7
8
9
250
ADVANCED ALGEBKA
N.
1
2 1 3
4
5
6
7
8
9
20
3010
3032
3054
3075
3096
3118
3139
3160
3181
3201
21
22
23.
21
25
26
27
28
29
3222
3424
3617
3802
3979
4150.
4314
4472
4624
3243
3444
3636
3820
3997
4166
4330
4487
4639
3263
3464
3655
3838
4014
4183
4346
4502
'4654
3284
3483
3674
3856
4031
4200
4362
4518
4669
3304
3502
3692
3874
4048
4216
4378
4633
4683
3324
3522
3711
3892
4065
4232
4393
4648
4698
3346
3541
3729
3909
4082
4249
4409
4564
4713
3366
3560
3747
3927
4099
4265
4426
4679
4728
3385
3579
3766
3945
4116
4281
4440
4594
4742
3404
3698
3784
3962
4133
4298
4456
4609
4757
30
4771
4786
4800
4814
4829
4843
4857
4871
4886
4900
31
32
33
34
35
36
37
38
39
4914
5051
5185
5316
5441
5563
5682
5798
5911
4928
5065
5198
5328
5453
5575
5694
5809
5922
4942
5079
5211
5340
6465
5587
5705
5821
5933
4955
5092
6224
6353
6478
5599
6717
5832
5944
4969
5105
5237
5366
6490
6611
6729
5843
6955
4983
6119
5250
5378
6502
6623
5740
5855
5966
4997
6132
5263
6391
6614
6635
5752
6866
6977
5011
6145
6276
5403
6627
5647
5763
5877
6988
5024
5159
5289
6416
5539
5W>8
5775
5888
6999
5038
6172
5302
5428
5551
6670
5786
5899
6010
40
6021
6031
6042
6063
6064
6075
6085
6096
6107
6117
41
42
43
44
45
46
47
48
49
6128
6232
6335
6435
6532
6628
6721
6812
6902
6138
6243
6345
6444
6542
6637
6730
6821
6911
6149
6253
6356
6454
6551
6646
6739
6830
6920
6160
6263
6365
6464
6561
6656
6749
6839
6928
6170
6274
6376
6474
6671
6665
6758
6848
6937
6180
6284
6385
6484
6580
6675
6767
6857
6946
6191
6294
6395
6493
6590
6684
6776
6866
6955
6201
6304
6405
6603
6599
6693
6785
6875
6964
6212
6314
6415
6613
6609
6702
6794
6884
6972
6222
6325
6425
6522
6618
6712
6803
6893
6981
SO
6990
^6998
7007
7016
7024
7033
7042
7050
7069
7067
51
52
53
54
55
56
57
58
59
7076
7160
7243-
7324
7404
7482
7559
7634
7709
7084
7168
7251
7332
7412
7490
7566
7642
7716
7093
7177
7259
7340
7419
7497
7574
7649
7723
7101
7185
7267
7348
7427
7505
7582
7657
7731
7110
7193
7275
7356
7435
7513
7689
7664
7738
7118
7202
7284
7364
7443
7620
7597
7672
7745
7126
7210
7292
7372
7451
7628
7604
7679
7752
7136
7218
7300
7380
7459
7536
7612
7686
7760
7143
7226
7308
7388
7466
7643
7619
7694
7767
7162
7235
7316
7396
7474
7551
7627
7701
7774
60
7782
7789
7796
7803
7810
7818
7825
7832
7839
7846
61
62
63
64
65
66
67
68
69
7853
7924
7993
8062
8129
8195
8261
8325
8388
7860
7931
8000
8069
8136
8202
8267
8331
8395
7868
7938
8007
8075
8142
8209
8274
8338
8401
7875
7945
8014
8082
8149
8215
8280
8344
8407
7882
7952
8021
8089
8166
8222
8287
8351
8414
7889
7959
8028
8096
8162
8228
8293
8357
8420
7896
7966
8035
8102
8169
8235
8299
8363
8426
7903
7973
8041
8109
8176
8241
8306
8370
8432
7910
7980
8048
8116
8182
8248
8312
8376
8439
,7917
7987
8055
8122
8189
8254
8319
8382
8446
70
8451
8467
8463
8470
8476
8482
8488
8494
8500
8506
N.
1
2
3
4
5
6
7
8
9
LOGARITHMS
251
N.
1
2
3
4
5
6
7
8
9
70
8451
8457
8463
8470
8476
8482
8488
8494
8600
8506
71
72
73
8513
8573
8633
8519
8579
8639
8525
8585
8645
8531
8591
8651
8537
8597
8657
8543
8603
8663
8549
8609
8669
8555
8615
8675
8561
8621
8681
8567
8627
8686
74
75
76
8692
8751
8808
8698
8756
8814
8704
8762
8820
8710
8768
8825
8716
8774
8831
8722
8779
8837
8727
8785
8842
8733
8791
8848
8739
8797
8854
8745
8802
8869
77
78
79
8865
8921
8976
8871
8927
8982
8876
8932
8987
8882
8938
8993
8887
8943
8998
8893
8949
9004
8899
8954
9009
8904
8960
9015
8910
8965
9020
8915
8971
9025
80
9031
9036
9042
9047
9053
9058
9063
9069
9074
9079
81
82
83
9085
9138
9191
9090
9143
9196
9096
9149
9201
9101
9154
9206
9106
9159
9212
9112
9165
9217
9117
9170
9222
9122
9175
9227
9128
9180
9232
9133
9186
9238
84
85
86
9243
9294
9345
9248
9299
9350
9253
9304
9355
9258
9309
9360
9263
9315
9365
9269
9320
9370
9274
9325
9375
9279
9330
9380
9284
9335
9385
9289
9340
9390
87
88
89
9395
9445
9494
9400
9450
9499
9405
9455
9S04
9410
9460
9509
9415
9465
9513
9420
9469
9518
9425
9474
9523
9430
9479
9528
9435
9484
9533
9440
9489
9538
90
9542
9547
9552
9557
9562
9666
9671
9576
9581
9586
91
92
93
9590
9638
9685
9595
9643
9689
9600
9647
9694
9605
9652
9699
9609
9657
9703
9614
9661
9708
9619
9666
9713
9624
9671
9717
9628
9675
, 9722
9633
9680
9727
94
95
96
9731
,9777
9823
9736
9782
9827
9741
9786
9832
9745
9791
9836
9750
9795
9841
9754
9800
9845
9759
9806
9850
9763
9809
9854
9768
9814
9859
9773
9818
9863
97
98
99
9868
9912
9956
9872
' 9917
9961
9877
9921
9965
9881
9926
9969
9886
9930
9974
9890
9934
9978
9894
9939
9983
9899
9943
9987
9903
9948
9991
9908
9952
9996
100
0000
0004
0009
0013
0017
0022
0026
0030
0035
0039
N.
1
2
3
4
5
6
7
8
9
216. Exponential equations. Equations in wHcli the variable
occurs only in the exponents may often be solved by the use of
tables of logarithms if one keeps in mind the fact that
log a' = x log a.
EXERCISES
Solve the following :
1. 10»-i = 4.
Solution : Taking the logarithm of both sides of the equation, we have
(a; -1) log 10 = log 4,
or since log 10 = 1, a; = log 4 + 1 = .6021 + 1 = 1.6021.
252 ADVANCED ALGEBRA
2. 4^-31' = 8,
2'= ■ By = 9.
Solution : Taking the logarithms of the equations, we have
a; log 4 + 2/ log 3 = log 8,
a; log 2 + 2^ log 8 = log 9,
or a;21og2 + 2/log3 = 31og2, (1)
a; log2 + y 3 log2 = 2 log 3.
Eliminate x.
a;21og2+ 2/log3 = 81og2
a:21og2 + y61og2 = 41og3
y (log3 - 6 log 2) = 3 log2 - 41og3
_ 31og2 - 41og3 _ 3 X .3010 - 4 x .4771
^~ log3-61og2 ~ .4771-6 X. 3010
.90.30 - 1.9084 _ - 1.0064 _ 1.0054
.4771 - 1.8060 ~ - 1.3289 ~ 1.3289*
Perform this division by logarithms.
log 1.0054 = 10.0023-10
log 1.3289= .1235
logy= 9.8788-10
y= .7565.
Substituting in (1),
a; = 3 log 2 - .7565 log 3 _ .9030 - .7565 x .4771
2 log 2 .6020
Compute .7565 x .4771 by logarithms.
log. 7565= 9.8788-10
log. 4771= 9.6786-10
log result = 19.5574 - 20
result = .3609.
.9030 - .3609 .5421
Hence x = =
.6020 .6020
log.5421 = 19.7341 -20
log. 6020= 9.7796-10
loga!= 9.9545-10
X = .9006.
3.©' = 2. 4. 4!- = 3. 5. 7" + '= 5.
6. 32=>^ + i = 5. 7. 4^-1 = 5^+1. 8. 22»+s-6"^-
-1 = 0.
' c" ■ d« = n. ' x + y = n.
LOGARITHMS 253
' x-y = i. ■ 3a; + 2y = 17.
,„ 3»^ ■ i" = 15552, -. V^^^. Va3»-i = a«,
• 4^ . &' = 128000. "• ^psTs . ^pFTi = fiio.
217. Compound interest. If $1600 is at the yearly interest
of 3%, the total interest for a year is $1500 ■ (0.03) = $46. The
total sum invested at the end of a year would be $1646.
Let, in general, P represent a sum of money in dollars.
Let r represent a yearly rate of interest.
Then P ■ r represents the yearly interest on P, and
p + p.j. = p(}. + l)
represents the total investment, principal and interest, at the
end of a year.
Similarly, P (r + 1) r is the second year's interest, and
P(r + !)?• + P(r + 1) = P(?-2 + /• + r + 1) = P(r + 1)2
is the total investment at the end of two years.
In general, A = P (r + 1)" (1)
is the total accumulation at the end of n years. If we know r, P,
and n, we can by (1) find A. If we take the logarithm of both
sides of the equation, we have
log A = log P + n log (r + 1),
^^ log^-logP ,
log(r- + l) ^ ^
Hence if we know A, P, and r, we can find n.
If the interest is computed semiannually, we have as interest
at the end of a half-year -P-k' "while the entire STun would be
P (- + 1 ) . Reasoning as above, we find that if the iaterest is com-
puted semiannually, the accumulation at the end of n years is
^=p(^ + l)". (3)
Similaxly, »= ^"S^-logP ^^^
2 log
(i-)
254 ADVANCED ALGEBRA
If the interest is computed k times a year, -we liave at tlie end
of n years
log 4 — logP ,.,
or n= —^ —2 — ^6)
k
log (I 4- 1)
EXERCISES
In sucli exercises as the following, four-plaoe tables are not sufficiently
exact to obtain perfect accuracy. In general, the longer the term of years
and the more frequent the compounding of interest, the greater the inaccuracy.
1. If 11600 is placed at S\% interest computed semiannually for 13 years,
to how much will it amount in that time ?
Solution : By formula (3), A = pi^ + l\ ".
F = 1600, r = .OSJ, n = 13.
/.07 \26 / 7 \26 /407\26
Hence ^ = 1600(— + 1) =1600(-i- + l) = 1600 ( — ) .
\4 / V400 / UOO/
log 1600= 3.2041 log 407 =2.6096
261og407 = 67.8496 ^^
71.0537 52192
26 log 400 = 67.6546 67.8496
logA= 3.3991 log400= 2.6021
A = $2507. 156128
52042
67.6546
2. After how long will |600 at 6% computed annually amount to $1000 ?
Solution : By formula (2) we have
^_ log^-logP
log(r + l)
A = 1000, P = 600, r = .06.
log 1000 - log600 3 - 2.7782 .2218 „ „^
n = — = = = = = 8.76 years.
log 1.06 .0253 .0263 ^
.76 year = .76 -.12 = 9.12 months.
.12 month = .12 ■ 30 = 3.6 days.
Thus n = 8 years 9 months 3.6 days.
LOGAEITHMS 255
In the following exercises the interest is computed annually unless the
contrary is stated.
3. To what will $3750 amount in 20 years if left at 5% interest ?
4. To what sum will |25,300 amount in 10 years if left at 5% interest
computed semiannually?
5. To what does $1000 amount in 10 years if left at 6% interest computed
(1) annually, (2) semiannually, (3) quarterly ?
6. A sum of money is left 22 years at 4% and amounts to $17,000. How
much was originally put at interest ?
7. What sum of money left at 4^% for 30 years amounts to $30,000 ?
8. What sum of money left 10 years at 4i% amounts to the same sum as
$8549 left 7 years at 5% ?
9. If a man left a certain sum 11 years at 4%, it would amount to $97 less
than if he had left the same sum 9 years at 5%. What was the sum ?
10. Which yields more, a, sum left 10 years at 4% or 4 years at 10% ?
What is the difference for $1000 ?
11. Two sums of money, $25,795 in all, are left 20 years at 4|%. The
difference in the sums to which they amount is $14,660. What were the sums ?
12. At what per cent interest must $15,000 be left in order to amount to
$60,000 in 32 years ?
13. At what per cent must $3333 be left so that in 24 years it will
amount to $10,000?
14. Two sums of which the second is double the first but is left at 2% less
interest amount in 36J years to equal sums. At what per cent interest was
each left ?
15. In how many years .will a sum double if left at 5% interest ?
16. In how many years will a sum double it left at 6% interest computed
semiannually ?
17. In how many years will a sum amount to ten times itself if left at
4% interest ?
18. In how many years will $17,000 left at 4J% interest amount to the
same as $7000 left at 5J% for 20 years ?
19. On July 1, 1850, the sum of $1000 was left at 4J% interest. When
paid back it amounted to $2222. When did this occur ?
20. Prove formulas (1), (3), and (5) by complete induction.
CHAPTEE XXI
CONTINUED FRACTIONS
218. Definitions. A fraction in the form
ff + --;,
where a, J, • • •, g', ■ ■ • are real numbers, is called a continued fraction.
We shall consider only those continued fractions in which the
numerators b, d, f, etc., are equal to unity and in which the
letters represent integers, as for example
„ -L. 1 -.^ ,111
Oi T , 1 written ch-\ — —
«2+l_ aa + ai + ai-A .
as+ 1_
a,i-\ ,
When the number of quotients a^, a^, »<, ■ ■ • is finite the frac-
tion is said to be terminating. When the fraction is not terminat-
ing it is infinite. We shall see that the character of the numbers
represented by terminating fractions differs widely from that of
the numbers represented by injEinite continued fractions. We shall
find, in fact, that any root of a linear equation in one variable,
i.e. any rational number, may be represented by a terminating
continued fraction, and conversely; furthermore, that any real
irrational root of a quadratic equa(tion may be represented by the
simplest type of infinite continued fractions, and conversely.
219. Terminating continued fractions. If we have a terminat-
ing continued fraction, where a^, a^, ■■■ are integers, it is evident
that by reducing to its simplest form we obtain a rational num-
ber. The converse is also true, as we can prove in the following
256
CONTINUED FRACTIONS 257
Theorem. Any rational number may he expressed as a ter-
minating fraction.
Let - represent a rational number. Divide a by h, and let a^
be the quotient and e (which must be less than J) the remainder.
Then (§ 26) ,
a c 1
- = a, + - = a, + -.
Divide b by e, letting a^ be the quotient and d (which must be
less than c) the remainder. Then
a ^1
da + «
C
Continuing this process, the maximum limit of the remainders
in the successive divisions becomes smaller as we go on, until
finally ^he remainder is zero. Hence the fraction is terminating.
It is noted that the successive quotients are the denominators in
the continued fraction.
EXERCISES
1. Convert the following into continued fractions
. (a) ttVt-
Solution: 247] 77 [0
77J247[3
231
16J77[4
64
13J16[1
13
3J13[4
12
1)313
3
The continued fraction is
77 _ 1 1 1 1 1
247 ~ 3 + 4 + 1 + 4 + S"
258 ADVANCED ALGEBRA
(") if- (c) iVt- (d) m-
(e) Hi- (f) m- . (g) //A-
w ms- (i) Ml- u) Hif
2. Express the following continued fractions as rational fractions.
2 + 3 ^ ' 4 + 7 ^ ' a + 6
(d)l I (e)l 1 1. (f)l 1 1.
K + a; ''' 1+2 + 3 ^ ' 3 + 4 + 5
(g) 1 1 1 1 1 (hi 1 i i - 1 -
^'2 + 4 + 2 + 4 + 2' ^ S + 2 + 3 + 1 + 2 + 3"
220. Convergents. The value obtained by taking only the first
n — 1 quotients in a continued fraction is called the nth. convergent
of the fraction.
Thus in the fraction
1+1111
2+3+2+6
1 is the first convergent,
1 3
1 + - = - is the second convergent,
l + -=l+- = — is the third convergent, etc.
3
■When there is no whole number preceding the fractional part of the continued
fraction the first convergent is zero. Thus in
111
2 + 3 + 5
^ is called the second convergent.
In the continued fraction
^1111
«! H — — — —
a-2 + as + «4 + «6 H
let -i , — , — , ... represent the successive convergents expressed
?i ?2 9s
as rational fractions.
Then for the first convergent we have
Oi = — , or^i = «!, qi = 1.
CONTINUED FRACTIONS 259
For the second convergent we have
1 a^Oi+l Pi
«! H = ="^> or J72 = flSaai + 1 = a,pi + 1,
(Jg <^2 2'2
S'a = a2 = a^qi.
Tor the third convergent we have
1 tta as (gggi + !)+«! ^3
% H , 1 = «i H TT = — ^ Ti = ~ '
^2 + 2^ a3ei2 + 1 <''zO-2 +1 ?3
da
or Pa=<h («^2% + 1) + «! = «3i'2 +Pi,
23 = «sa2 + 1 = azqi + ?i-
This indicates that the form of the mth convergent is
Pn ^ a'nPu-l+Pn-2
This is in fact the case, as we proceed to show by complete
induction.
We have abeady established form (1) for ?i = 2 and Ji = 3.
We assume it for n = m, and will show that its validity for
n = m + 1 follows. The (m + l)th convergent differs from the
mth only in the fact that a„ H appears in the continued
fraction in place of a„. In (1) replace n by m, and a„ by
a^ H ) and we have
(1)
Pi
2m +
(a„, + - )i'™_l+i'„-2
+ 1 _ V "'m + U
^ ('*'» + ;;: — )2'».-i + ?».-2
^ (fl'm + lO'm + i)Pm-1. + a„ + lj>„-2
(««+,!«'». + 1)2».-X + a'm + lSm-2
_ O'm+lid-mPm-l + Pm-2) + Pn,-1
«.« + l(«,»?m-l + Sm-2)+Sm-l
_ <^m+lPm+Pm-l
O'm + lim + ?™_1
which is form (1).
260 ADVANCED ALGEBRA
EXERCISES
1. Express the following as continued fractions, and find the convergents.
(a) a-
Solution : By the method already explained, we find that
30_1 11111
41 "1 + 2 + 1 + 2 + 1 + 2'
Here ai = 0, 02 = 1, as = 2, 04 = 1, a^ = 2, Oe = 1, a^ = 2.
The first convergent is evidently 0, the second is 1, and the third is
1 2
mi. j: ^i, . ■ Pi atPs +P2 1-2+1 3
The fourth convergent is — = --^^ — — = = - •
S4 aiSa + g2 1-3 + 1 4
The fifth convergent is ^ = ^^P±±Pl = ^-1+1 = 1.
55 asqi + qs 2-4 + 8 11
™, . „ ^ ■ Pe aePs +Pt 1-8 + 3 11
The sixth convergent is — = -^^-^ — ^ = = —
go (Hqs -f gi 1-11 + 4 15
™ , ^ ■ P7 aiPs +P5 2-11 + 8 30
The seventh convergent is — = -^^ — — = ■ — = —
g? flvge + 55 2-15 + 11 41
(b) m- (c) If- (d) t¥i- (e) ^A- (f) m-
(g) i¥i- W f'A'ti- (i) i¥r- U) m"!-
2. Find the value of the following by finding the successive convergents.
,,11111 ,,,11111
(a) - - - - -■ (D) - - - - -■
^'2+1+2+1+2 ^.3+2+3+2+3
111111 (dl-i-i-i
^°' 2 +,3 + 1 + 1 + 3 + 2' ^'3 + 3 + 3 + 3 + 3 + 3'
111111 /f\ 111111
^®' 5 + 3 + 1 + 1 + 3 + 5' ^'1 + 3 + 5 + 6 + 3 + 1'
^^' (X - 1) + X + (X + 1) ^ ' X + X + X
221. Recurring continued fractions. We have seen that every
terminating continued fraction represents a rational number, and
conversely. We now discuss the character of the numbers repre-
sented by the simplest infinite continued fractions. A recurring
continued fraction is one in w^hich from a certain point on, a
group of denominators is repeated in the same order.
CONTINUED FRACTIONS 261
111111
"" 3 + 2 + 3 + 2 + 3+2 + '"'
111111
1 + 2 + 3 + H-2 + 3+'"
are recuiring continued fractions if the denominators are assumed to repeat
indefinitely as indicated.
That a repeating continued fraction actually represents a num-
ber we shall establish in § 223. Unless this fact is proven, one
runs the risk of dealing with symbols which have no meaning.
If for certain continued fractions the successive convergents
increase without limit, or take on erratic values that approach no
limit, it is important to discover the fact. All the fractions that
we discuss actually represent numbers, as we shall see.
We shall consider only continued fractions in which every
denominator has a positive sign.
Theorem. Every recurring continued fraction is the root of a
quadratic equation.
T ^ * ■ + 111111
Let, for instance, x = - - - - - - ••■.
' ■' a+ b+ o + a+ b + +
Evidently the part of the fraction after the first denominator c
may be represented by x, and we have thus virtually the termi-
nating fraction
111
X =
a -\- b -\- -\- X
The second convergent is —
0/
The third convergent is
. ^ ^Ps
ab + 1 2-3
The fourth convergent, or x, gives us
_Pi _ O'iPs+Pi ^ (e +x)b + l
qi O'iii + 22 (« + x) {ab +l)+a
Simplifying, we get
{ab -1- l)a;^ 4- [c{ab + 1)+ a -b'\x - be -1 = 0,
262 ADVANCED ALGEBRA
which is a quadratic equation whose root is x, the value of the
continued fraction.
Since this equation has a negative number for its constant term
it has one positive and one negative root. The continued fraction
must represent the positive root, since we assume that the letters
a, b, c represent positive integers. The quadratic equation whose
root is a recurring continued fraction with positive denomina^
tors will always have one positive and one negative root. The
equation will be quadratic, however, whatever the signs of the
denominators may be.
The proof may be extended to the case where there are any
number of recurring denominators or any niimber of denominators
before the recurrence sets in. Since every real irrational root
of a quadratic equation is a surd, our result is equivalent to the
statement that every recurring continued fraction may be expressed
as a surd.
EXERCISES
Of what quadratic equations are the following roots ? Express the con-
tinued fraction as a surd.
1 1 I 11
■ 2+3+2+3+""
Solution : Let x = -
2 + 3 + a:
Then x = l 1 - ^ + ^
2 + - 6 + 2a;+l
3 + a;
or 2 a:2 + 6 a; - 3 = 0.
Solving this equation, we get
-3+V15 _3-Vl5
Xi = OT Xi =
2 2
Since xi is negative, xi must he the surd that is represented by the con-
tinued fraction.
1111 -3+VT5
Thus
2+3+2+3+ 2
allll... 3IIII
■ 1 + 2 + 1 + 2 + ""' ■ 3 + 2 + 3 + 2 + "
CONTINUED FRACTIONS 263
^111111 gllll
■ 1 + 2 + 3 + 1 + 2 + 3 + '"" ■ 3 + 1 + 3 + 1 + '"'
6.2 + 1 I I I ;... 7.3 + 1 1 1 ....
2+1+2+1+ ^3+4+5+
Hint. Ifa; = 2 + ^ i ..., g^ 1 1 1 1 ...
1 ^j+^+ +3 + 4 + 3 + 4 +
then «-2=-^-^... g 1 1 1 1 1 1
and « = 2 + l 1 - ■ 1+2 + 3 + 1 + 2 + 3+ ' '
^2 + l + (a;-2)
10. i i 1 1 1 1 ...
2+1+2+2+1+2+
222. Expression of a surd as a recurring continued' fraction.
This is the converse of the problem discussed in the last sec-
tion, and shows that recurring continued fractions and quadratic
equations are related in the same intimate way that terminat-
ing fractions and rational numbers (i.e. the roots of linear equa-
tions) are connected. We seek to express an irrational number,
as, for instance, Vz, as a continued fraction. This we may do
as follows.
Since 1 is the largest integer in V2 we may write
V2 = 1 +- ( V2 - 1) = 1 + (-^^^^.
Rationalizing the numerator, we have
V2-I-I
Since 2 is the largest integer in V2 -f 1 we have
V2 = H- A= r = l+ ^
24-(V2-l) o I (V2-I)
EationaKzing the numerator V2 — 1, we have
^V2+l ^ + 2+(V2-l)
264 ADVANCED ALGEBRA
By continuing this process we continually get the denomi-
nator 2. Thus 1 1
This process consists of the successive application of two opera-
tions, and affords the
EuLB. Express the surd as the sum of two numbers the first
of which is the largest integer that it contains.
Rationalize the numerator of the fraction whose numerator is
the second of these numbers. Repeat these operations until a
recurrence of denominato7's is observed.
This process may be apphed to any surd, and a continued frac-
tion which is recurring will always be obtained. We shall con-
tent ourselves with a statement of this fact without proof.
If the surd is of the form a — Vi, a continued fraction may be
derived for -|- Vft and its sign changed. Since the real roots of
any quadratic equation x^ + 2 a^x + aj = are surds of the form
a±^/b, where a and h are integers, it appears that the roots of
any such equation may be expressed as recurring continued frac-
tions. It can be shown that the real roots of the general quad-
ratic equation aox^ + a^x -|- aj = may also be so expressed.
EXERCISES
1. Express the following siirds as recurrent continued fractions,
(a) 2 + VS.
Solution :
2-fV3 = 3 + (V3-l) = 3+ -' ^ (1)
3 ■- 1 9
= 3-f 4=— '- = 3 + -
Vs-t-i Vs + i
= 3 + -^~ = 3 + -i = 3 + L
•^ + 1 1 V3+1 V3-I
2 2 "^2
= ^ + j| 8-1 =^ + I| J_ = 3 + ^ 1
2(V3 + 1) V3-H 2-KV3-I)
X
CONTINUED FRACTIONS 265
But since Vs — 1 is tlie same number tliat we have in (1), this fraction
repeats from this point on, and we have
2+V3 = ;
3 +
1111
1+2+1+2+"
••
(b) V5. •
(c) vTf.
(d) V65.
(e) V47.
(f) Vii.
(g) V23.
(h) V34.
(i) Vi9.
(i) V62.
(k) V79.
(1) V98.
(m) V88.
(n) V22.
(0) V45.
(p) V59.
(q) VlOi,
(r) 7+vn.
(s)
8-
-V3.
(t)
3-V23.
2. Express as a continued fraction the roots of the following equations,
(a) a;2-7a;-3 = 0. (b) a;2 + 2a; - 6 = 0.
(c) a;2 + 3a;-8 = 0. (d) s2-4a; - 4 = 0.
223. Properties of convergents. The law of formation of con-
vergents given in § 220 is valid whether the continued fraction
is terminating or infinite. We should expect that in the case of
an infinite fraction the successive convergents would give us an
increasingly close approximation to the value of "the fraction.
This is indeed the fact, as we shall see.
Theokem. The difference between the nth and (n + l)st con-
vergents is ^
We prove this theorem by complete induction.
Let the continued fraction be
«! H — — —
»2 + «3 + <»4 +
Then the first and second convergents are respectively
Then ^_£i^ '(«.<^x + l) _ 1.
?! Si «8 <h
(1)
266, ADVANCED ALGEBRA
Since qi = 1, g'a = a-i,
we hare ^^^ - -^^ = tll)!^ f or w = 1.
We assume that the theorem holds for n = m, that is,
Pm + l _ Pm _ Pn,qm+l-qv^v, + l ^ (-!)"■+ \
S'm+l ?»! ^mS'm + l ymym + l
We must prove that it holds for m = m + 1.
Now since ^"'+'' — ^^i+i _ i'^+ig'^+g ~ J'm+agm+i^
our theorem reduces to proving that the numerator
i'™+i2'm+2-i'».+2?m+i =(-!)'"+'• (2)
In the left-hand member of (2) set
«m + 2?». + l + 2m = 2m + 2) (1)> § 220
and a,n, + 2Pm + l+Pm=Pm + S-
Then Pm+l(<^m+22m + l + Sm) - (o-m + iPm+l +Pm)Sm + l
= Pm + l<lm - Pm2m + i.= -{Pm2m+\ — Pm+lird
= _(_l)-+i=(_l)'»+2. by (1)
CoROLLAEY I. The difference between the successive convergents
of €b continued fraction with positive denominators approaches
zero as a limit.
Since q„ — a„q„_i + q„_2, evidently q„ increases without limit
when n is increased, siace to obtain q„ we add together positive
numbers neither one of which can vanish. .
Thus we can find a value of n large enough so that — > and
1 ^»
hence > will be smaller than any assigned number, which
?»!Z» + i 1
is another way of stating that as n increases approaches
zero as a limit.
CONTINUED FKACTIONS 267
CoEOLLAEY II. The even convergents decrease, while the odd
convergents increase, as n increases.
We must show that
Pm+2 _Prn
im + 2 im
is negative or positive according as m is even or odd. Adding
and subtracting - ^"'"^' ) we have
Pm + i Pm (Pm + i _ PjH+l \ i ( Pm+l A
_ (-l)-+^ ^ (-1)" + ^
2'?ii + 2?m+l Sm + lSm
By Corollary I, the denominator of the first fraction exceeds
that of the second. Hence when m is odd the sum in the last
member of the equation is positive, and when m is even the sum
is negative.
We now see that any recurring fraction of the type considered
in § 221 actually represents a number in the sense of § 74. We
have seen that the successive odd convergents continually increase,
while the even convergents continually decrease, until the differ-
ence between a pair of them is very small. Such sequences of
numbers we have seen (§ 73 ff.) define real numbers.
224. Limit of error. We are now in a position to state a maxi-
mum value for the error made in taking any convergent of a con-
tinued fraction for the fraction itself.
Theorem. The maximum limit of error in taking the nth
convergent for the continued fraction is less than
Since by the theorem of the last section the value of the frac-
tion is between any pair of consecutive convergents, it must
differ from either of these convergents by less than they differ
from each other, that is, by less than
?»?» + !
268 ADVANCED ALGEBRA
EXERCISES
rind a convergent that differs by less than .001 from each of the following :
1. V6.
Solution :
V6 = 2 + (V6-2) = 2+ ^~^ =2 + .
V6 + 2 V6 + 2
U(^-^)-'-l-
2
= 2 + ^,/ V6 + 2 „\^2+^ , V6-2
= 2 + - 6-4 =2 + - 1 =2 + - 1
^ + 2(V6 + 2) ^'^V^T^ ^ + i + (V6-2)
Since the last surd repeats the one in the first equation we have
V6 = 2 + l 1 1 1 ....
^2+4+2+4+
Pi 2 . 1)2 5 . ps 4.5 + 2 22 .
qi~l' q^~2' gs 4.2+l~9'
Pi 2.22 + 5 49. j)5_4.49 + 22_
218
94 ~ 2 .9 + 2 20 ■ 85 4-20 + 9
89
^ ^ - ^ -001
qtqs 20.89 1780 ''
Since
we see by § 224 that f J satisfies the condition of the problem.
2. V7. 3. Vie. 4. Vs. 5. Vi9.
6. V35. 7. V32. 8. Vel. 9. V55.
10. 3+V2. 11. V99. 12. Vli. 13. Vis.
14. The number TT has the value 3. 14159. Find by the method of continued
fractions a series of convergents the last of which differs from this value by
less than .0001.
CHAPTEE XXII
INEQUALITIES
225. General theorems. We say that a is greater than b when
OS — 5 is positive. If a — b is negative, then a is less than b.
Thus any positive number or zero is greater than any negative
number. As we distinguished between identities and equations
of condition in § 53, so in this discussion we observe that some
statements of inequality are true for any real value of the letters,
while others hold for particular values only. The former class
may be called unconditional inequalities, the latter conditional.
Thus a^ > _ 1 ia true for any real value of a and is unconditional, while
a; — 1 > 2 only when x is greater than 3 and is consequently conditional.
The two inequalities a> b, c> d are said to have the same
sense. Similarly, a < b, a < d have the same sense. The inequal-
ities a> b, <. d have a different sense.
Theorem I. Awi/ positive number may be added to, subtracted
from, or multiplied by both numbers of an inequality without
affecting the sense of the inequality.
Let a^b, that is, let a — 5 = A, where k is a positive number.
If m is a positive number, evidently
a ± TO — (6 ± m) = A;,
or a ± TO > 5 ± m.
Similarly, ma — mb = mlc,
or ma > mb.
The other statements of the theorem are proved similarly.
CoEOLLAKY. Terms may be transposed from one side of an
inequality to the other as in the ease of equations.
270 ADVANCED ALGEBRA
Let a> b + c.
Subtract c from Taotli sides of the inequality and we obtain by
Theorem I
a — c> b.
Theorem II. If the signs of both sides of an inequality are
changed, the sense of the inequality must be reversed, that is, the
> sign must be changed to <, or conversely.
Let a>b, that is, let a — 6 = A, where A; is a positive number.
Then —a + h = —k,
or (— a) — (— S) = — A,
that is, by definition, — a< — b.
EXERCISES
Prove that the following "identities are true for all real positive values of
the letters.
1. a2 + 62>2a6.
Solution : (a — 6)^ is alvcays positive.
Thus a2 - 2 a6 + 62 = a^ + 62 - 2 a6 is positive.
That is, a^ + 62 > 2 db.
2. 3(a3 + 6')>a26 + a62.
3. a2 + 62 + c2 > a6 + ac + he.
4. (6 + c) (c + o) (a + 6) > 8 abc.
5. (a + 5 + c) (o2 + 62 + c2) > 9a6c.
6. 62c2 + c2a2 + a262 > a6c (o + 6 '+ c).
7. 3 (aP + 6' + c8) > (a + 6 + c) (a6 + 5c + ca).
8. V(x + «i)2 +(31 + yiY < Vx2 + 2/2 + Va;i2 + y^a.
9. If o2 + 62 = 1, 0:2 + 2/2 = 1^ prove that ax + by< 1.
10. (a + 6 - c)2 + (a + c - 6)2 + (6 + c - a)2 > a6 + 6c + ca.
1 1. Show that the sum of any positive number (except 1) and its reciprocal
is greater than 2.
12. Prove that the arithmetical mean of two unequal positive numbers
always exceeds their geometrical mean.
INEQUALITIES 271
226. Conditional linear inequalities. If we wish to find the
values of x for which.
ax + b < B, (1)
where a, b, ajid c are numbers and a is positive, we may find
such values by carrying out a process similar to that of solving a
linear equation in one variable.
By the corollary, § 225, we have from (1)
ax < c — b.
By Theorem I, § 225, x < ^-^^—
•' a
227. Conditional quadratic inequalities. We have abeady
shown in § 116 that the quadratic expression ax^ + te + c is posi-
tive or negative, when the equation
ax" + bx + c=zO (1)
has imaginary or equal roots, according as a is positive or negS/-
tive. If the equation has distinct real roots, the expression is
positive or negative for values between those roots according as
a is negative or positive. This we may express in tabular form
as follows, for all values of x excepting the roots of (1), for
which of course the expression vanishes.
a
t^-iac
c(3? + bx + c
+
-orO
Always +
—
-orO
Always —
+
+
— for X between roots, + for other values
-
+
+ for X between roots, — for other values
This enables us to answer immediately questions like the
following :
Example. Eor what values of a; is — 2 x" + a; > — 3 ? By the corollary, § 225,
this is equivalent to the question, For what value of a; is —2x" + x + S>0?
Here 6^ _ 4 (jc = 1 + 24 = 25 is positive. The roots of the equation
— 2 a;2 + a; + 3 = are a; = — 1, a; = f . Thus by our table this expression
is positive tor all values of x between — 1 and |.
272 ADVANCED ALGEBRA
EXERCISES
For what values of x are the following inequalities valid ?
1. 2a;-3>0. 2. 4z-7>l.
3. -s-l>7. 4. -3x + 8<3.
_ 9a; 4 , . 2a; , 3^4
3 7 3 4 5
7. .12a; + .3 < 1.3. 8. 3-4x>2.
9. 3<5x-2. 10. .8<^-l.
11. a;!" - 8x + 22 > 6. 12. x^ + 3x - 2 > 1.
13. 2a;2-3a;>5. 14. -3x2-4x>8.
15. 2a;i'-4a;< -2. 16. 3a;2 _ 9a;> - 6.
17. -3x2 + 2s<2. 18. -x2 + 6a;>9.
19. 5a;2-8x<l. 20. a;2<x-l.
21. 3x2>3x-3. 22. 3s>2xa-4.
CHAPTEE XXIII
VARIATION
228. General principles. The number x is said to vary directly
as the number y when the ratio of a; to y is constant. This we
symbolize by
. xxy,OT:- = k, (1)
where ^ is a constant.
Thus if a man walks at a uniform speed, the distance that he
goes varies directly as the time. If the length of the altitude of
a triangle is given, the area of the triangle varies directly as the
base. The volume of a sphere varies directly as the cube of
its radius.
The number x is said to vary inversely as the number y when
X varies directly as the reciprocal of y. Thus x varies inversely
as y when ^
a; oc -) or - = ccy = A, (2)
y
where A is a constant. Thus the speed of a horse might vary
inversely as the weight of his load. The length of time to. do
a piece of work might vary inversely as the number of laborers
employed.
The intensity of a light varies inversely as the square of the
distance of" the light from the point of observation. If I repre-
sents the intensity of light and d the distance of the Eght from
the point of observation, we have
where A; is a constant;
d?
27.8''
274 ADVANCED ALGEBKA
The number x is said to vary jointly as y and a when it varies
directly as the product of y and «. Thus x varies jointly as y
and z when
xocyz,oi: — = k, (4)
where A: is a constant.
Thus a man's wages might vary jointly as the number of days
and the number of hours per day that he worked.
The number x is said to vary directly as y and inversely as s
when it vanes directly with -• Thus the force of the attraction
of gravitation between two bodies varies directly as their masses
and inversely as the squares of their distances. If m represents
the masses of two bodies, d their distance, and O the force of
their attraction due to gravity, then
m Gd^
Gcc-^> or = k. Co)
d^ m ^ ^
^ EXERCISES
Ijjitfvaries inversely as the square of 6, and if a = 2 when 6 = 3, what
is the value of a when 5 is 18 ?
l^olution: By (3), alfl = k.
We can determine k by substituting a = 2, 6 = 8.
2 • 9 = fc.
18 = fc.
Then a ■ (18)2 = 18,
or a = A-
2. The volume of a sphere varies as the cube of its radius. A sphere of
radius 1 has a volume 4.19. What is the volume of a sphere of radius 3 ?
Solation: Let V represent the volume and r the radius of the sphere.
Then by (1),
Determine k by substituting,
Then
V
= ifc.
4.19
1
= fc.
k--
= 4.19.
V
(3)"
4.19.
F= 113.13.
VARIATION 275
3. If a;^ a a; + y, and x = 1 when y = 1, find x when j/ = 8.
4. The area of a circle varies as the square of the radius. If a circle of
radius 1 has an area 3.14, find the area of a circle whose radius is 21.
5. Find the volume of a sphere whose radius is .2.
Hint. See exercise 2.
6. The volume of a circular cylinder varies jointly with the altitude and
the square of the radius of the base. A cylinder whose altitude and radius
are each 1 has a volume of 3.14. Find the volume of a cylinder whose
altitude is 15 and whose radius is 3.
7. The weight of a body of a given material varies directly with its
volume. If a sphere of radius 1 inch weighs } of a pound, how much would
a ball of the same material weigh whose radius is 16 inches ?
8. The distance fallen by an object starting from rest varies as the square
of the time of falling. If a body falls 16 feet in 1 second, how far will it
fall in 6 seconds 1
9. A body falls from the top to the bottom of a clifl in 3J seconds. How
high is the cliH ?
10. A triangle varies in area jointly as its base and altitude. The area
of a triangle whose base and altitude are each 1 is J. What is the area of a
triangle whose base is 16 and altitude 7?
11. If 6 men do a piece of work in 10 days, how long will it take 5 men
to do it?
12. If 3 men working 8 hours a day can finish a piece of work in 10 days,
how many days' will 8 men require if they work 9 hours a day ?
13. An object is 30 feet from a light. To what point must it be moved in
order to receive (a) half as much light, (b) three times as much light ?
14. The weights of objects near the earth vary inversely as the squares
of their distances from the center of the earth. The radius of the earth is
about 4000 miles. If an object weighs 150 pounds on the surface of the
earth, how much would it weigh 6000 miles distant from the center ?
CHAPTER XXrV
PROBABILITY
229. Illustration. If a bag contains 3 white balls and 4 blact
balls, and 1 ball is taken out at random, what is the chance that
the ball drawn will be white ?
This question we may answer as follows : There are 7 balls in
the bag and we are as likely to get one as another. Thus a ball
may be drawn in 7 different ways. Of these 7 possible ways 3
will produce a white ball. Thus the chance that the ball drawn
will be white is 3 to 7, or |. The chance that a black ball will
"be drawn is ^.
230. General statement. It is plain that we may generalize
this illustration as follows : If an event may happen in p ways
and fail in q ways, each way being equally probable, the chance
or probability that it will happen in one of the p ways is
^. ' (1)
The chance that it wiU fail is
i' + 2'
(2)
The sum of the chances of the ^vent's happening and failing
is 1, as we observe by adding (1) and (2).
The odds in favor of the event are the ratio of the chance of
happening to the chance of failure. In this case the odds in
favor are
-• (3)
The odds agannst the event are — .
V
276
PEOBABILITX 277
EXERCISES
1. If the cliance of an event's happening is ^, what aie the odds in its
favor ?
Solution! By(l), _£_ = JL.
Hence Wp = p + q,
or 9p = q,
or = n 1 which by (3) are the odds in favor.
q 9
2. Trom a pack of 52 cards 3 are missing. What is the chance that they
are all of a particular suit t
Solution : The number of combinations of 62 cards taken 3 at a time is
52 . 51 . 50 „, .
CJ2, s = — This represents p + q. The number of combinations of the
18 • 12 • 11
13 cards of any one suit taken 3 at a time is cis, 3 = ■ „ • This repre-
sents p,
13 ■ 12 ■ 11
Thus P ^ 1-2.3 ^ 13-12.11 ^ 11 ^ 11
p + q 62 • 61 • 50 52. 51- 50 17-60 860"
1.-2.3
3. What is the chance, of throwing one and only one 6 in a single throw
of two dice ?
Solution : There are 36 possible ways for the two dice to fall. This repre-
sents p + q. Since a throw of two sixes is excluded there are 5 throws in
which each die would be a 6, that is, 10 in all in which a 6 appears. This
represents p.
p 10 5
Thus
p + q 36 18
4. A bag contains 8 white and ^12 black balls. What is the chance that a
ball drawn shall be (a) white, (b) black ?
5. A bag contains 4 red, 8 black, and 12 white balls. What is the chance
that a ball drawn shall be (a) red, (b) white, (c) not black?
6. In the previous problem, if 3 balls are drawn, what is the chance that
(a) all are black, (b) 2 red and 1 white ?
7. What is the chance of throwing neither a 3 nor a 4 in a single throw
of one die ?
8. What is the chance in drawing a card from a pack that it be (a) an
ace, (b) a diamond, (c) a face card ?
278 ADVANCED ALGEBEA
9. Three cards are missing from a pack. What is the chance that they
are (a) of one color, (b) face cards, (c) aces ?
10. A coin is tossed twice. What is the chance that heads will fall
once ?
H. The chance that an event will happen is f. What are the odds in its
favor ?
12. The odds against the occurrence of an event are |. What is the chance
of its happening ?
13. What is the chance of throwing 10 with a single throw of two dice ?
14. A squad of 10 men stand in line. What is the chance that A and B
are next each other ?
15. What is the chance that in a game of whist a player has 6 trumps ?
16. What is the chance that in a game of whist a player holds 4 aces ?
CHAPTEE XXV
SCALES OF NOTATION
231, General statement. The ordinary numbers with which
we are acquainted are expressed by means of powers of 10. Thus
263 = 210'' + 6-10i + 3.
This is the common scale of notation, and 10 is called the radix
of the scale.
In a similar manner a number might be expressed in any scale
with any radix other than 10. If we take 6 as the radix, we shall
have as a number in this scale, for instance,
543 = 6-6=' + 4-6 + 3.
In this scale we need only and five digits to express every
positive integer.
In general, if r is the radix of a scale of notation, any positive
integer N will be denoted in this scale as follows :
N = ttor" + ai?-"-' + a^r"-^ H \- a„. (1)
THEOREM. Any positive integer may he expressed in a scale
of notation of radix r.
Suppose we have a positive integer N. Let r" be the highest
power of r that is contained in N. Then
N = doT-" + Ni,
where Ni is less than r". Suppose that on dividing Ny by r"~' we
obtain ^^ ^ ^^^„_i ^ ^^^
where N^ is less than r"-\
Then N = a^-f" + ajr"-' + N^.
Proceeding in this manner we obtain finally
N = do''" + ai**""' H h a„,
where the a's are positive integers less than r, or perhaps zeros.
279
280 ADVANCED ALGEBRA
One observes that the symbol 10 indicates the radix in any
system. In this general scale we need a and ) — 1 digits to
express every possible number.
232. Fundamental operations. In the four fundamental operar
tions in the common scale we carry and borrow 10 in computing.
In computing in a scale of radix 6, for instance, we should carry
and borrow 6. If the radix were r, we should carry or borrow r.
Thus let r = 6. Then 4 + 5 = 1-6 +3 = 13. Similarly, 5- 3 = 2-6 +.3= 23.
This is precisely analogous to our computation in the common scale, where, for
instance, we would have 9 + 8 = 1 • 10 + 7 = 17, or 6 ■ 7 = 4 ■ 10 + 2 = 42.
EXERCISES
^ Perform the following operations.
■ 1. 2361 + 4263 + 2140 ; r = 7.
2361
4253
2140
12114
In this process, since 3 + 1=4 and is less than the radix, there is nothing to carry.
The next column gives 6 + 5 + 4= 15= 2-7 + 1, hence we write down 1 and carry 2. The
next column gives 3 + 2 + 1 + 2= 8= 1-7 + 1, hence we write down 1 and carry 1. Finally
we get 2 + 4 + 2 + 1=9 = 1-7 + 2, hence we write 12.
2. 4602 - 3714 ; r = 8.
4602
3714
666
Since we cannot take 4 from 2 we borrow one from the next place. Since the radix
is 8 this amounts to 8 units in the first place. We then subtract 4 from 8 + 2, which
leaves 6. In borrowing 1 from that digit is really reduced to 7 and the preceding digit
to 5 ; then subtracting 1 from 7 we get 6. Since we cannot take 7 from 5 we borrow 8
again and take 7 from 5 + 8= 13, which leaves 6. Since 1 has been borrowed from the 4
we see the subtraction is complete since 3-3=0.
3. 4321.432; r = 5.
4321
432
14142
24013
33334
4143222
In multiplying by 2 we have nothing to carry until we multiply 3 by 2. This gives
6=1-5 + 1. Hence we put down 1 and carry 1 to the product of 2 and 4. The addition of
the partial products is carried out as in exercise 1.
SCALES or NOTATION 281
4. 32130 -- 43 ; r = 6.
43J 32130 |430
300
213
213
00
In making an estimate for the iirst figure in tlie quotient we divide 32 by 4, keeping
in mind tliat for tliis purpose 32=3-6 + 2. Tlius 4 is contained in 20 just 5 times, but
since our entire divisor is 43 we take 4 as the iirst iigure in the quotient. The multipli-
cations are of course performed as in exercise 3, excepting that here 6 is the radix.
5. 4361 + 2635 + 5542 ; r = 7. 6. 6344 - 3456 ; r = 7.
7. 2340-4101; r = 5. 8. 6435-35; r = 7.
9. 2003455 ^ 403 ; »• = 6. 10. 344032 -^ 321^ r = 5.
11. 534401 - 443524; r = 6. 12. 425 + 254 + 542 + 452 ; r = 6.
233. Change of scale. If we have a number in the scale of
radix r, we may find the expression for that number in the com-
mon scale by writing the number in form (1), § 231, and carrying
put the indicated operations.
Example. Convert 4635, where r = 7, into the ordinary scale.
4635 = 4- 73 -f6- 72 + 3-7-1-5
= 4-343 + 6-49 + 3-7+5
= 1692.
If we have a number in the common scale, we may express it
in the scale with radix r as follows : If the number is N, we have
to determine the integers ao, a-i, •■•, a„ in the expression
~ N = aa'f + ayV^-'^ + ■■■-+ a;^_■^r + a„. (1)
Divide (1) by r. We have
- = a„r«- 1 -f ai?'"-^ -f a„_- + ^ = iV' -I- ^ ;
that is, the remainder a„ of this division is the last digit in the
expression desired.
' Divide iV' by r and we obtain
— = N" = af,i^-' +- ffiir"-^ H -|- ^^^ ;
r - r
282 ADVANCED ALGEBRA
that is, the remainder from this division is the next to the last
digit in the desired expression. Proceeding in this way we obtain
all the digits a„, a„_i, • ■ ■, a^, a^.
Example. Express 37496 in the scale with radix 7.
7) 37496
7 ) 6356 remainder 4
1 ) 765 remainder 1
7 )109 remainder 2
7 )15 remainder 4
7 )2 remainder 1
remainder 2
The number in scale r = 7 is 214214.
To change a number from any scale r-j to any other scale rj,
we may first change the number to the scale of 10 and then by
the process just given to the scale r^. The process indicated in
the preceding example may be employed directly to change from
any scale to any other, provided the division is carried out in the
scale in which the number is given. One of these methods may
be used to check the other.
Example, Change 34503 from scale r = 6 to one in which r = 9.
34503 = 3- 6* + 4- 68 +5- 62 + 3 = 4935 in scale of 10.
9 )4935
9) 548 remainder 3
9 )60 remainder 8
9)6 remainder 6
remainder 6
Thus 34503 in scale of 6 becomes 6683 in scale of 9.
Check; 9) 34503
9) 2812 remainder 3 ^" carrying out this division it must
V • A a ^^ kept in mind that the dividends are in
922^ remainder « ^^^^ ^^ e^ ^^^^^ ^^^ remainders are to be
9)10 remainder 6 in scale of 9.
remainder 6
234. Fractions. In the ordinary notation we express fractional
numbers by digits following the decimal point. This notation
may also be used in a scale with any radix.
SCALES OF NOTATION 283
Thus the expression .6421 stands for
10 10^ 10» ^ 10*
in the common scale.
In the scale with radix r it stands for
'-AAA-
lyt /jit'' ipO /yC
The process of changing the scale for fractions is performed
in accordance with the same principles as are employed in the
change of scale for integers. The following examples suffice to
illustrate it. ■/
Example 1. Express .5421 in the scale Of 6 as a decimal fraction.
.5421 = ^ + 1+^+1
' 6 62 68 6*
_ 6-6° + 4.68 + 2.6 + l _ 1237 _
"" 6* ~1296~"
Example 2. Express .439 as a fraction for radix 6.
Let .439 = ? + l + -^ + l + ....
6 62 63 6*
Multiplying by 6, 2.634 = a + ^ + | + 1 + ■ • • .
Thus a = 2 and we have .634 = - H 1 1 .
6 62 63
c d
Multiplying by 6, 3.804 = & + - + — + ■•-.
6 62
Thus 6 = 3 and we have .804 = - H h • • ..
6 62
Multiplying by 6, 4.824 = c + - + ••■ .
6
Thus c = 4 and we have .824 = - ^ .
6
Multiplying by 5, 4.944 = d-\ .
Thus d = 4.
The fraction in scale of radix 6 is then .2344 ....
28^ ADVANCED ALGEBRA
EXERCISES
1. Express the following as decimal fractions.
(a) .374; r = 8. (b) .4352 ; r = 6.
(c) .2231 ; r = 4. (d) .2001 ; r = 3.
2. Express the decimal fraction .296 as a radix fraction for r = 5.
3. Express the decimal fraction .3405 as a radix fraction for r = 6.
4. Express — as a radix fraction for r = 4.
128
5. Express — as a radix fraction for r = 5.
^ 626
6. In what scale is 42 expressed as 1120 ?
Solution : We seek r where
rS + r2 + 2 r = 42.
This is. equivalent to "finding a positive integral root of the equatioai
r^ + r2 + 2 r - 42 = 0.
By synthetic division, , •
1 + 1+ 2-42|2
+ 2+ 6 + 16
+ 3 + 8-26
1 + 1 + 2 - 42[3
+ 3+12+42
. . + 4 + 14
Thus 3 is the value sought
Check : 3^ + 3= + 2 ■ 3 = 27 + 9 + 6 = 42.
7. In what scale is 2704 denoted by 20304 ?
8. In what scale is 256 denoted by 10000 ?
9. In what scale is .1664 denoted by .0404?
10. Show that 1331 is, a. perfect cube. [
235. Duodecimals. We may apply some of the foregoing
processes to mensuration.; If we take one foot as a unit and the
radix as 12, we may express distances in the so-called duodecimal
notation. Thus 2 ft. 6 in. is represented in the duodecimal scale
by 2.6. Since in a scale of radix r we need r —1 symbols, we
will let 10 = ^ and 11 = e. Thus 21 ft. 10 in. would be expressed
in duodecimal notation as 19,*. We may now find areas and vol-
umes in this notation much more readily than by the usual method
of converting all distances to inches.
SCALES OF NOTATION 285
Example. Multiply 8 ft. 3 in. by 3 ft. 10 in. "We multiply 8-3 by 3.«
8.3 in the duodecimal scale. To convert the result to square
^■t feet and square inches yre must keep in mind that 27. 76
= 2 • 12 + 7 + ^ + tI, = 31 sq. ft. 90 sq. in., since 144
square inches equal one square foot.
6t6
209
27.76
This example suggests the following method of multiplying
distances :
KuLE. Express the distances in duodecimal notation with the
foot as a unit.
Multiply in the scale for which r = 12.
In the product change the part on the left of the point from
duodecimal to decimal scale.
Multiply the digit following the point ly W, and add to the
last figure to obtain the square inches in the result.
EXERCISES
1. Multiply the following :
(a) 13 ft. 4 in. by 67 ft. 11 in.
Solution: 11.4
67.e
1028
794
568
635.68 = 905 sq. ft. 80 sq. in.
(b) 10 ft. 6 in. by 12 ft. 2 in.
(o) 8 ft. 4 in. by 11 ft. 11 in.
(d) 23 ft. 6 in. by 47 ft. 8 in. '
(e) 41 ft. 6 in. by 36 ft. 1 in.
2. What is the area of a room 16 ft' 2 in. by 10 ft. 3 in. ?
3. What is the area of a walk 60 ft. 6 in. by 3 ft. 3 in. ?
4. What is the area of a city lot 52 ft 6 in. by 153 ft 7 in. ?
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